Trigonometry and the Complex Plane

This section covers:

In certain physics and engineering applications, it’s easier to perform certain computations with complex numbers (numbers that give a negative result when squared), rather than with real numbers (“normal” numbers).

In this section, we’ll learn how to convert Rectangular Form coordinates and equations to Polar (Trig) Complex Form, in order to perform these computations.

Review of Complex Numbers

We learned that complex numbers exist so we can do certain computations in math, even though conceptually the numbers aren’t “real”. We learned about them here in the Imaginary (Non-Real) and Complex Numbers section.

To work with complex numbers and trig, we need to learn about how they can be represented on a coordinate system (complex plane), with the “\(x\)”-axis being the real part of the point or coordinate, and the “\(y\)”-axis being the imaginary part of the point. For example, the representation of the complex number \(z=x+yi\) would be \((x,y)\) in the complex plane.

Here is a visual representation of a point in the complex plane, with it’s magnitude. The magnitude (sometimes called modulus) of a complex number is like the hypotenuse of a triangle, with lines drawn to the \(x\) (real) and \(y\) (imaginary) coordinates as the sides of the triangles. Thus, the magnitude of \((4,3)\) or \(4+3i\) is \(\sqrt{{{{4}^{2}}+{{3}^{3}}}}=\sqrt{{25}}=5\).

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This form of a trig number is sometimes written as \(z=a+bi\) instead of \(z=x+yi\)  (in this case, \(z=4+3i\)). And remember that \(i=\sqrt{{-1}}\), although we won’t really need to use this in this section.

You can also put complex numbers in your graphing calculator, and even perform complex conversions on the graphing calculator, as shown after each section.

Polar (Trig) Form of a Complex Number

Again, we can write the rectangular form of a complex number in the form \(z=x+yi\), or more commonly, \(z=a+bi\).

We can also write this in a trig polar form, where \(x=r\cos \theta \) and \(y=r\sin \theta \). Notice we can use the abbreviation cis or CIS (for cos plus i sin), since the angles measurements (\(\theta \)) are the same.

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“\(r\)” is called the magnitude or modulus of \(z\), like we saw earlier, and is sometimes written as \(\left| z \right|\). The angle \(\theta \) is called the argument.

\(\displaystyle z=x+yi=\left( {r\cos \theta } \right)+\left( {r\sin \theta } \right)i\) which can be abbreviated to \(\displaystyle z=r\cdot \text{cis}\left( \theta \right)\) or \(\displaystyle z=r\cdot \text{cis}\theta \)

Note that \(\displaystyle \left( {r\cos \theta } \right)+\left( {r\sin \theta } \right)i\)  actually turns out to be the equivalent of \(r{{e}^{{\theta \,i}}}\) (Euler’s equation); in fact, we’ll see when we use the calculator to check answers. (We won’t prove this here, or use it any more).

Here is a visual example; note that we get the cosine and sine of 45° from the Unit Circle, where we learned about here in the Angles and the Unit Circle section:

Converting Complex Rectangular Form to Polar Form

When converting from Rectangular to Polar Form, to get \(r\) and \(\theta \), we have to use the same equations we did here in the Polar Coordinates, Equations and Graphs section here:

\(\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}\,\,\,\,\,\,\,\,\,\text{(this will be positive)}\)

\(\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)\,\,\,\,\,\,\text{(check for correct quadrant)}\)

Note that when using the calculator to get \({{\tan }^{{-1}}}\), you’ll have to add the following degrees or radians when your point is in the following quadrants. This is because the \({{\tan }^{{-1}}}\) function on your calculator only give you answers back in the interval \(\displaystyle \left( {-\frac{\pi }{2},\frac{\pi }{2}} \right)\).

Here are some examples (I’ve left the argument in degrees). Remember that \(x\) is the real number and \(y\) is the imaginary number (coefficient of “\(i\)”).

Rectangular Complex Form Convert to Polar Complex Form
\(-1+\sqrt{3}i\) \(\begin{array}{c}r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{{\left( {\sqrt{3}} \right)}}^{2}}}}=\sqrt{{1+3}}=\sqrt{4}=2\\\theta ={{\tan }^{{-1}}}\left( {\frac{{\sqrt{3}}}{{-1}}} \right)=120{}^\circ \text{ (2nd quadrant)}\\2\left( {\cos 120{}^\circ +i\sin 120{}^\circ } \right)=2\,\text{cis}\left( {120{}^\circ } \right)\end{array}\)     
\(-1-i\) \(\begin{array}{c}r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{{\left( {-1} \right)}}^{2}}}}=\sqrt{{1+1}}=\sqrt{2}\\\theta ={{\tan }^{{-1}}}\left( {\frac{{-1}}{{-1}}} \right)=225{}^\circ \text{ (3rd quadrant)}\\\sqrt{2}\left( {\cos 225{}^\circ +i\sin 225{}^\circ } \right)=\sqrt{2}\,\text{cis}\left( {225{}^\circ } \right)\end{array}\)             
\(-3\) \(\displaystyle \begin{array}{c}r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{{{{\left( {-3} \right)}}^{2}}+0{}^{{^{2}}}}}=\sqrt{9}=3\\\theta ={{\tan }^{{-1}}}\left( {\frac{0}{{-3}}} \right)=180\text{ (between 2nd }\\\text{ and 3rd quadrants)}\\3\left( {\cos 180{}^\circ +i\sin 180{}^\circ } \right)=3\,\text{cis}\left( {180{}^\circ } \right)\end{array}\)             
\(i\) \(\displaystyle \begin{array}{c}r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{0{}^{2}+{{1}^{2}}}}=\sqrt{1}=1\\\theta ={{\tan }^{{-1}}}\left( {\frac{1}{0}} \right)=\text{undefined (between 1st }\\\text{ and 2nd quadrants: 90}{}^\circ \text{)}\\\cos 90{}^\circ +i\sin 90{}^\circ =\text{cis}\left( {90{}^\circ } \right)\end{array}\)             
\(2-5i\) \(\displaystyle \begin{array}{c}r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{{{2}^{2}}+{{{\left( {-5} \right)}}^{2}}}}=\sqrt{{4+25}}=\sqrt{{29}}\\\theta ={{\tan }^{{-1}}}\left( {\frac{{-5}}{2}} \right)=291.8{}^\circ \text{ (calculator gives us -68}\text{.2}{}^\circ \text{, but we }\\\text{ need to add 360}{}^\circ \text{, since in 4th quadrant)}\\\sqrt{{29}}\left( {\cos 291.8{}^\circ +i\sin 291.8{}^\circ } \right)=\sqrt{{29}}\,\text{cis}\left( {291.8{}^\circ } \right)\end{array}\)             

 

How to Check with Graphing Calculator

You can use the graphing calculator to convert back and forth between polar to rectangular coordinates (like we saw in the Polar Coordinates, Equations and Graphs section), and even find complex powers and roots.

For rectangular-polar conversions (in either DEGREE or RADIAN mode), use 2nd APPS (ANGLE) 5 [R-Pr(] and 2nd APPS (ANGLE) 6 [R-PƟ(], typing in the rectangular form without the “\(i\)” and using a comma between the real and imaginary parts. Note that you have to solve for the \(r\) (which may be in decimal and not exact form) and \(\theta \) separately.

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You can also use MATH CMPLX → 7 (Polar), after typing in the rectangular \(a+bi\) form. Note that the answer is in the form \(r{{e}^{{\,\theta \,i}}}\); when converting back this way, you’ll need to put your calculator in mode RADIAN and “re^(θi)” (and convert the angle to radians) and enter it this way.

Let’s try converting \(-1-i\) to Polar. Notice that you may have to add 360° if you get a negative angle.


Converting Polar Form to Rectangular Form

Now let’s write the polar complex numbers in rectangular form. To do this, you just get the sin and cos of the angle, and then push through (distribute) the number in front (the “\(r\)”). Again, this is similar to what we did in the Polar Coordinates, Equations and Graphs section here much easier going this way!

(These match up with the conversions above, and some are in radian mode instead of degree mode).

Polar Complex Form Convert to Rectangular Complex Form
\(2\,\text{cis}\left( {120{}^\circ } \right)\) \(\begin{align}2\,\text{cis}\left( {120{}^\circ } \right)&=2\left( {\cos 120+i\sin 120} \right)\\&=2\left( {-\frac{1}{2}+i\cdot \frac{{\sqrt{3}}}{2}} \right)=-1+\sqrt{3}i\end{align}\)
\(\displaystyle \sqrt{2}\,\text{cis}\left( {\frac{{5\pi }}{4}} \right)\) \(\begin{align}\sqrt{2}\,\text{cis}\left( {\frac{{5\pi }}{4}} \right)&=\sqrt{2}\left( {\cos \frac{{5\pi }}{4}+i\sin \frac{{5\pi }}{4}} \right)\\&=\sqrt{2}\left( {-\frac{{\sqrt{2}}}{2}+i\cdot -\frac{{\sqrt{2}}}{2}} \right)=-1-i\end{align}\)
\(3\,\text{cis}\left( \pi \right)\) \(\begin{align}3\,\text{cis}\left( \pi \right)&=3\left( {\cos \pi +i\sin \pi } \right)\\&=3\left( {-1+i\cdot 0} \right)=-3\end{align}\)
\(\text{cis}\left( {90{}^\circ } \right)\) \(\begin{align}\text{cis}\left( {90{}^\circ } \right)&=\cos 90{}^\circ +i\sin 90{}^\circ \\&=0+i\cdot 1=i\end{align}\)
\(\displaystyle \sqrt{{29}}\,\text{cis}\left( {291.8{}^\circ } \right)\) \(\begin{align}\sqrt{{29}}\,\text{cis}\left( {291.8{}^\circ } \right)&=\sqrt{{29}}\left( {\cos 291.8{}^\circ +i\sin 291.8{}^\circ } \right)\\&\approx 2-5i\end{align}\)

How to Check with Graphing Calculator

For polar-rectangular conversions (in either degrees or radians mode), use 2nd APPS (ANGLE) 7 [P-Rx(] and 2nd APPS (ANGLE) 8 [P-Ry(], typing in the rectangular form using a comma between \(r\) and \(\theta \). Note that you have to solve for the \(x\) and \(y\) separately, and also note again that you won’t get the answers with the roots in them (you’ll get decimals that aren’t “exact”). You also won’t get the “\(i\)” in the \(y\) part of the answer.

You can also use MATH CMPLX → 6 (Rect), but you most first put the calculator in RADIAN and re^(θi) modes. Use the polar \(r{{e}^{{\,\theta \,i}}}\) form, making sure that the degrees value is converted to radians.

Let’s try converting \(\displaystyle \sqrt{2}\,\text{cis}\left( {\frac{{5\pi }}{4}} \right)\) to Rectangular:

Products and Quotients of Complex Numbers in Polar Form

You can use the polar form of a complex number to find products and quotients of complex numbers; you’ll basically want to memorize these formulas.

Just remember that when you multiply two polar complex numbers, you multiply the magnitudes (numbers in front), but add the angle measurements. When you divide two polar complex numbers, you divide the magnitudes (numbers in front), but subtract the angle measurements. Pretty weird, huh?

\(\displaystyle {{z}_{1}}={{r}_{1}}\left( {\cos {{\theta }_{1}}+i\sin {{\theta }_{1}}} \right)=\,{{r}_{1}}\,\text{cis}\left( {{{\theta }_{1}}} \right)\,\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,\,\,{{z}_{2}}=\,\,{{r}_{2}}\left( {\cos {{\theta }_{2}}+i\sin {{\theta }_{2}}} \right)={{r}_{2}}\,\text{cis}\left( {{{\theta }_{2}}} \right)\)

 

\(\displaystyle \,\,\,\,\,\,\,\,\text{Product:}\,\,\,\,{{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left[ {\cos \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)} \right]={{r}_{1}}{{r}_{2}}\,\text{cis}\left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)\)

\(\displaystyle \text{Quotient:}\,\,\,\,\frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left[ {\cos \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)} \right]=\frac{{{{r}_{1}}}}{{{{r}_{2}}}}\,\text{cis}\left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)\)

Here are some examples; both types of notation are used for complex polar numbers. We’ll leave our answers in Polar Form between 0 and 360° or 0 and \(2\pi \):

Complex Numbers Products and Quotients
\(z=2\,\text{cis}\left( {20{}^\circ } \right);\,\,\,\,\,w=4\text{cis}\left( {40{}^\circ } \right)\)

 

\(\displaystyle \text{Find }zw\text{ and }\frac{z}{w}\)

\(\begin{align}zw&={{r}_{1}}{{r}_{2}}\left[ {\cos \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)} \right]\\&=2\cdot 4\cdot \text{cis}\left( {20+40} \right)=8\,\text{cis}\left( {\text{60}{}^\circ } \right)\\\frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left[ {\cos \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)} \right]\\&=\frac{2}{4}\cdot \text{cis}\left( {20-40} \right)=\frac{1}{2}\text{cis}\left( {-\text{20}{}^\circ } \right)=\frac{1}{2}\text{cis}\left( {34\text{0}{}^\circ } \right)\end{align}\)
\(\displaystyle z=4\,\text{cis}\left( {\frac{\pi }{8}} \right);\,\,\,\,\,w=3\text{cis}\left( {\frac{{9\pi }}{{16}}} \right)\)

 

\(\displaystyle \text{Find }zw\text{ and }\frac{z}{w}\)

\(\begin{align}zw&={{r}_{1}}{{r}_{2}}\left[ {\cos \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)} \right]\\&=4\cdot 3\cdot \text{cis}\left( {\frac{\pi }{8}+\frac{{9\pi }}{{16}}} \right)=12\,\text{cis}\left( {\frac{{11\pi }}{{16}}} \right)\\\frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left[ {\cos \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)} \right]\\&=\frac{4}{3}\cdot \text{cis}\left( {\frac{\pi }{8}-\frac{{9\pi }}{{16}}} \right)=\frac{4}{3}\text{cis}\left( {\frac{{-7\pi }}{{16}}} \right)=\frac{4}{3}\text{cis}\left( {\frac{{25\pi }}{{16}}} \right)\end{align}\)
\(z=1-i;\,\,\,\,\,w=1+\sqrt{3}i\)

 

\(\displaystyle \text{Find }zw\text{ and }\frac{z}{w}\)

(degrees)

\(\begin{array}{c}z=1-i:\\\\r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{{{1}^{2}}+{{{\left( {-1} \right)}}^{2}}}}=\sqrt{2}\\\theta ={{\tan }^{{-1}}}\left( {\frac{{-1}}{1}} \right)=315{}^\circ \\\text{(4th quadrant)}\\\sqrt{2}\,\text{cis}\left( {315{}^\circ } \right)\end{array}\)

\(\begin{array}{c}z=1+\sqrt{3}i:\\\\r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{{{1}^{2}}+{{{\left( {\sqrt{3}} \right)}}^{2}}}}=2\\\theta ={{\tan }^{{-1}}}\left( {\frac{{\sqrt{3}}}{1}} \right)=60{}^\circ \\\text{ (1st quadrant)}\\2\,\text{cis}\left( {60{}^\circ } \right)\end{array}\)
\(\displaystyle \begin{align}zw&={{r}_{1}}{{r}_{2}}\left[ {\cos \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}+{{\theta }_{2}}} \right)} \right]\\&=\sqrt{2}\cdot 2\,\text{cis}\left( {315+60} \right)=2\sqrt{2}\,\text{cis}\left( {375{}^\circ } \right)=2\sqrt{2}\,\text{cis}\left( {15{}^\circ } \right)\\\frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left[ {\cos \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)+i\sin \left( {{{\theta }_{1}}-{{\theta }_{2}}} \right)} \right]\\&=\frac{{\sqrt{2}}}{2}\cdot \text{cis}\left( {315-60} \right)=\frac{{\sqrt{2}}}{2}\text{cis}\left( {255{}^\circ } \right)\end{align}\)

How to Check with Graphing Calculator

To multiply and divide complex numbers, you can put these in the calculator in Polar Form using re^(θi) (RADIAN mode only), or Rectangular Form in either DEGREE or RADIAN mode. Let’s try \(\displaystyle 4\,\text{cis}\left( {\frac{\pi }{8}} \right)\cdot 3\,\text{cis}\left( {\frac{{9\pi }}{{16}}} \right)\) (in mode RADIAN), and \(\displaystyle \frac{{\left( {1-i} \right)}}{{\left( {1+\sqrt{3}i} \right)}}\) (in mode DEGREE):

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Note that we might have to add 360° to a negative angle to get the correct angle. And we can always convert the product or quotient back to Rectangular by using MATH CMPLX →Rect.

De Moivre’s Theorem: Powers of Complex Numbers

De Moivre’s Theorem (named after the French mathematician Abraham De Moivre), is a formula for raising a complex number to a power (greater than or equal to 1).

Just remember that when you raise a complex number to a power, you raise the magnitude to that power, but multiply the angle measurements. (Later we’ll see that when you take the root of a complex number, you take the root of the magnitude, but divide the angle measurements).

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See how it’s similar to multiplying and dividing complex numbers?

Here is De Moivre’s Theorem:

\(\displaystyle {{z}^{n}}={{r}^{n}}\left[ {\cos \left( {n\theta } \right)+i\sin \left( {n\theta } \right)} \right]\,={{r}^{n}}\left[ {\text{cis}\left( {n\theta } \right)} \right]\)

Here are some examples; both types of notation are used for complex polar numbers. We’ll write our answers in Standard or Rectangular Form \(a+bi\):

Complex Numbers Finding Powers using De Moivre’s Theorem
\({{\left[ {2\,\text{cis}\left( {80{}^\circ } \right)} \right]}^{{3}}}\) \(\displaystyle {{\left[ {2\,\text{cis}\left( {80{}^\circ } \right)} \right]}^{{\,3}}}={{2}^{3}}\,\text{cis}\left( {3\cdot 80{}^\circ } \right)=8\,\text{cis}\left( {\text{240}{}^\circ } \right)=8\left( {-\frac{1}{2}-\frac{{\sqrt{3}}}{2}i} \right)=-4-4\sqrt{3}i\)
\(\displaystyle {{\left[ {\sqrt{3}\,\text{cos}\left( {\frac{{7\pi }}{{18}}} \right)+i\sin \left( {\frac{{7\pi }}{{18}}} \right)} \right]}^{{6}}}\) \(\displaystyle \begin{align}{{\left[ {\sqrt{3}\,\text{cos}\left( {\frac{{7\pi }}{{18}}} \right)+i\sin \left( {\frac{{7\pi }}{{18}}} \right)} \right]}^{{\,6}}}&={{\left( {\sqrt{3}} \right)}^{6}}\text{cis}\left( {6\cdot \frac{{7\pi }}{{18}}} \right)={{3}^{{\frac{6}{2}}}}\text{cis}\left( {\frac{{42\pi }}{{18}}} \right)\\&=27\,\text{cis}\left( {\frac{{42\pi }}{{18}}-\frac{{36\pi }}{{18}}} \right)=27\,\text{cis}\left( {\frac{\pi }{3}} \right)\\&=27\left( {\frac{1}{2}+\frac{{\sqrt{3}}}{2}i} \right)=\frac{{27}}{2}+\frac{{27\sqrt{3}}}{2}i\end{align}\)
\({{\left( {1-i} \right)}^{7}}\)         Convert to Polar first:              \(\begin{array}{c}z=1-i:\\r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{{{1}^{2}}+{{{\left( {-1} \right)}}^{2}}}}=\sqrt{2}\\\theta ={{\tan }^{{-1}}}\left( {\frac{{-1}}{1}} \right)=315{}^\circ \text{(4th quadrant)}\\\sqrt{2}\text{cis}\left( {315{}^\circ } \right)\end{array}\)
\(\displaystyle \begin{align}{{\left[ {\sqrt{2}\text{cis}\left( {315{}^\circ } \right)} \right]}^{{\,7}}}&={{\left( {\sqrt{2}} \right)}^{7}}\text{cis}\left( {7\cdot 315{}^\circ } \right)={{2}^{{\frac{7}{2}}}}\text{cis}\left( {2205{}^\circ } \right)={{2}^{3}}\cdot {{2}^{{\frac{1}{2}}}}\text{cis}\left( {2205{}^\circ -360{}^\circ \cdot 6} \right)\\&=8\sqrt{2}\,\text{cis}\left( {45{}^\circ } \right)=8\sqrt{2}\left( {\frac{{\sqrt{2}}}{2}+\frac{{\sqrt{2}}}{2}i} \right)=8+8i\end{align}\)

How to Check with Graphing Calculator

To raise complex numbers to powers (De Moivre’s Theorem) using the calculator, again you can put these in the calculator using re^(θi) in RADIAN mode; let’s do these problems on the calculator: Find \({{\left[ {2\,\text{cis}\left( {80{}^\circ } \right)} \right]}^{{\,3}}}\) and find \({{\left( {1-i} \right)}^{7}}\).

Remember again that you have to use radians to input the 80° into the re^(θi) format; multiply by \(\pi \) and divide by 180 (and don’t forget the “\(i\)” in the exponent). Then, if we want the re^(θi) in degrees, we can change the mode to DEGREES, and hit “2nd – (ans) ENTER. For that problem, we get \(-4-4\sqrt{3}i\), like we got above. Note also that we can convert the complex numbers back to Rectangular by using MATH CMPLX →Rect.

Roots of Complex Numbers

To get roots of complex numbers, we do the opposite of raising them to a power; we take the \(n\)th root of the magnitude, and then divide the angle measurements by \(n\).

The only thing that’s a little tricky is there are typically many roots for a complex number, so we have to find all of these by the following formula, with \(k\) going from 0 to \((n-1)\):

\(\displaystyle \begin{align}\sqrt[n]{z}&=\sqrt[n]{r}\left[ {\cos \left( {\frac{\theta }{n}+\frac{{2\pi k}}{n}} \right)+i\sin \left( {\frac{\theta }{n}+\frac{{2\pi k}}{n}} \right)} \right],\,\,\,\,k=0,\,\,1,\,\,2,\,…,n-1\\&=\sqrt[n]{r}\,\text{cis}\left( {\frac{\theta }{n}+\frac{{2\pi k}}{n}} \right),\,\,\,\,k=0,\,\,1,\,\,2,\,…,n-1\end{align}\)

This looks really complicated, so let’s go through some examples (and get answers in either degrees or radians, per indicated). Let’s first work with roots with either the real or imaginary part of the complex number (but not both). For those roots that end up with special values (values on the Unit Circle), put the answers in \(a+bi\) form; for non-special values, leave in polar form.

You might also note that the roots of complex numbers are evenly spaced on the unit circle, all with the same magnitude.

Note that if a problem asks for a root of “unity”, this is just 1 \((1+0i)\).

Complex Numbers Finding Complex Roots
\(\sqrt[3]{{-27}}\)

 

Radians

  Convert  \(z=-27\)  to Polar first:

\(\displaystyle r=\sqrt{{{{{\left( {-27} \right)}}^{2}}+{{0}^{2}}}}=27;\,\,\,\,\,\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{0}{{-27}}} \right)=\pi :\,\,\,\, 27\text{cis}\left( \pi \right)\)       

\(\displaystyle \begin{array}{c}{{\left[ {27\text{cis}\left( \pi \right)} \right]}^{{\,\frac{1}{3}}}}={{27}^{{\frac{1}{3}}}}\text{cis}\left( {\frac{{\pi +2\pi k}}{3}} \right)\\=3\,\text{cis}\left( {\frac{{\pi +2\pi k}}{3}} \right);\,\,\,k=0,1,2\end{array}\)            \(\displaystyle \begin{array}{l}k=0:\,\,\,\,3\,\text{cis}\left( {\frac{{\pi +0}}{3}} \right)=3\,\text{cis}\left( {\frac{\pi }{3}} \right)=3\left( {\frac{1}{2}+\frac{{\sqrt{3}}}{2}i} \right)=\frac{3}{2}+\frac{{3\sqrt{3}}}{2}i\\k=1:\,\,\,\,3\,\text{cis}\left( {\frac{{\pi +2\pi }}{3}} \right)=3\,\text{cis}\left( \pi \right)=3\left( {-1+0i} \right)=-3\\k=2:\,\,\,\,3\,\text{cis}\left( {\frac{{\pi +4\pi }}{3}} \right)=3\,\text{cis}\left( {\frac{{5\pi }}{3}} \right)=3\left( {\frac{1}{2}-\frac{{\sqrt{3}}}{2}i} \right)=\frac{3}{2}-\frac{{3\sqrt{3}}}{2}i\end{array}\)
\(\sqrt[3]{{-64i}}\)

 

Degrees

Convert  \(z=64i\)  to Polar first:

\(\displaystyle r=\sqrt{{{{0}^{2}}+{{{\left( {-64} \right)}}^{2}}+}}=64;\,\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{{-64}}{0}} \right)\,\,\,\text{(und)}=\text{270}{}^\circ :\,\, 64\,\text{cis}\left( {\text{270}{}^\circ } \right)\)       

\(\begin{array}{c}{{\left[ {64\text{cis}\left( {\text{270}{}^\circ } \right)} \right]}^{{\frac{1}{3}}}}={{64}^{{\frac{1}{3}}}}\text{cis}\left( {\frac{{270+360k}}{3}} \right)\\=4\text{cis}\left( {\frac{{270+360k}}{3}} \right);\,\,\,\,\,k=0,1,2\end{array}\)            \(\begin{array}{l}k=0:\,\,\,\,4\,\text{cis}\left( {90{}^\circ } \right)=4\left( {0+i} \right)=\,\,4i\\k=1:\,\,\,\,4\,\text{cis}\left( {210{}^\circ } \right)=4\left( {-\frac{{\sqrt{3}}}{2}-\frac{1}{2}i} \right)=-2\sqrt{3}-2i\\k=2:\,\,\,\,4\,\text{cis}\left( {330{}^\circ } \right)=4\left( {\frac{{\sqrt{3}}}{2}-\frac{1}{2}i} \right)=2\sqrt{3}-2i\end{array}\)
Fifth Root of Unity

\((\sqrt[5]{1})\)

 

Degrees

Convert  \(z=1\)  to Polar first:

\(\displaystyle r=\sqrt{{{{1}^{2}}+{{0}^{2}}}}=1;\,\,\,\,\,\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{0}{1}} \right)=1:\,\,\,\, \text{cis}\left( 0 \right)\)       

\(\displaystyle \begin{array}{c}{{\left[ {1\text{cis}\left( 0 \right)} \right]}^{{\,\frac{1}{5}}}}={{1}^{{\frac{1}{5}}}}\text{cis}\left( {\frac{{0+360k}}{5}} \right)\\=1\text{cis}\left( {\frac{{360k}}{5}} \right);\,\,\,\,\,k=0,1,2,3,4\end{array}\)          \(\begin{array}{l}k=0:\,\,\,\,\text{cis}\left( {\frac{{360\left( 0 \right)}}{5}} \right)=\text{cis}\left( {0{}^\circ } \right)\\k=1:\,\,\,\,\text{cis}\left( {\frac{{360\left( 1 \right)}}{5}} \right)=\text{cis}\left( {72{}^\circ } \right)\\k=2:\,\,\,\,\text{cis}\left( {\frac{{360\left( 2 \right)}}{5}} \right)=\text{cis}\left( {144{}^\circ } \right)\end{array}\)      \(\begin{array}{l}k=3:\,\,\,\,\text{cis}\left( {\frac{{360\left( 3 \right)}}{5}} \right)=\text{cis}\left( {216{}^\circ } \right)\\k=4:\,\,\,\,\text{cis}\left( {\frac{{360\left( 4 \right)}}{5}} \right)=\text{cis}\left( {288{}^\circ } \right)\end{array}\)

 

Here are some more examples; we’ll keep all answers in trig form (radians). Notice how we want to find common denominators when adding the fractions with \(2\pi \) in them. Also note with the second problem how we turned \(\sqrt{{32}}\) into \(\sqrt{{{{2}^{5}}}}\) into \({{\left( {\sqrt{2}} \right)}^{5}}\) so we could simplify when we take the fifth root.

Complex Numbers Finding Complex Roots

Find fourth roots of

 

\(\sqrt{3}+i\)

 

  Convert \(\sqrt{3}+i\) to Polar first:

\(\displaystyle r=\sqrt{{{{{\left( {\sqrt{3}} \right)}}^{2}}+{{1}^{2}}}}=\sqrt{4}=2;\,\,\,\,\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{1}{{\sqrt{3}}}} \right)=\frac{\pi }{6}:\,\,\,\, 2\,\text{cis}\left( {\frac{\pi }{6}} \right)\)         

\(\displaystyle \begin{array}{c}{{\left[ {2\,\text{cis}\left( {\frac{\pi }{6}} \right)} \right]}^{{\,\frac{1}{4}}}}={{2}^{{\frac{1}{4}}}}\text{cis}\left( {\frac{{\frac{\pi }{6}+2\pi k}}{4}} \right)\\=\sqrt[4]{2}\,\text{cis}\left( {\frac{\pi }{{24}}+\frac{{2\pi k}}{4}} \right);\,\,\,\,k=0,1,2,3\\=\sqrt[4]{2}\,\text{cis}\left( {\frac{\pi }{{24}}+\frac{{12\pi k}}{{24}}} \right);\,\,\,\,k=0,1,2,3\end{array}\)              \(\displaystyle \begin{array}{l}k=0:\,\,\,\,\sqrt[4]{2}\,\text{cis}\left( {\frac{\pi }{{24}}+\frac{{12\pi \cdot 0}}{{24}}} \right)=\sqrt[4]{2}\,\text{cis}\left( {\frac{\pi }{{24}}} \right)\\k=1:\,\,\,\,\sqrt[4]{2}\,\text{cis}\left( {\frac{\pi }{{24}}+\frac{{12\pi \cdot 1}}{{24}}} \right)=\sqrt[4]{2}\,\text{cis}\left( {\frac{{13\pi }}{{24}}} \right)\\k=2:\,\,\,\,\sqrt[4]{2}\,\text{cis}\left( {\frac{\pi }{{24}}+\frac{{12\pi \cdot 2}}{{24}}} \right)=\sqrt[4]{2}\,\text{cis}\left( {\frac{{25\pi }}{{24}}} \right)\\k=3:\,\,\,\,\sqrt[4]{2}\,\text{cis}\left( {\frac{\pi }{{24}}+\frac{{12\pi \cdot 3}}{{24}}} \right)=\sqrt[4]{2}\,\text{cis}\left( {\frac{{37\pi }}{{24}}} \right)\end{array}\)

Solve:

 

\({{z}^{5}}=4-4i\)

 

(This is the same as taking the fifth roots of \(4-4i\))

 

Convert \(4-4i\) to Polar:

\(\displaystyle r=\sqrt{{{{4}^{2}}+{{{\left( {-4} \right)}}^{2}}}}=\sqrt{{32}}={{\left( {\sqrt{2}} \right)}^{5}};\,\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{{-4}}{4}} \right)=\frac{{7\pi }}{4}:\,\, {{\left( {\sqrt{2}} \right)}^{5}}\text{cis}\left( {\frac{{7\pi }}{4}} \right)\)       

\(\require {cancel} \displaystyle \begin{array}{c}{{\left[ {{{{\left( {\sqrt{2}} \right)}}^{5}}\text{cis}\left( {\frac{{7\pi }}{4}} \right)} \right]}^{{\,\frac{1}{5}}}}={{\left( {{{{\left( {\sqrt{2}} \right)}}^{{\cancel{5}}}}} \right)}^{{\frac{1}{{\cancel{5}}}}}}\text{cis}\left( {\frac{{\frac{{7\pi }}{4}+2\pi k}}{5}} \right)\\\,\,=\sqrt{2}\text{cis}\,\left( {\frac{{7\pi }}{{20}}+\frac{{2\pi k}}{5}} \right);\,\,\,\,k=0,1,2,3,4\\=\sqrt{2}\text{cis}\,\left( {\frac{{7\pi }}{{20}}+\frac{{8\pi k}}{{20}}} \right);\,\,\,\,k=0,1,2,3,4\end{array}\)            \(\begin{array}{l}k=0:\,\,\,\,\sqrt{2}\,\text{cis}\left( {\frac{{7\pi }}{{20}}+\frac{{8\pi \cdot 0}}{{20}}} \right)=\sqrt{2}\,\text{cis}\left( {\frac{{7\pi }}{{20}}} \right)\\k=1:\,\,\,\,\sqrt{2}\,\text{cis}\left( {\frac{{7\pi }}{{20}}+\frac{{8\pi \cdot 1}}{{20}}} \right)=\sqrt{2}\,\text{cis}\left( {\frac{{3\pi }}{{4}}} \right)\\k=2:\,\,\,\,\sqrt{2}\,\text{cis}\left( {\frac{{7\pi }}{{20}}+\frac{{8\pi \cdot 2}}{{20}}} \right)=\sqrt{2}\,\text{cis}\left( {\frac{{23\pi }}{{20}}} \right)\\k=3:\,\,\,\,\sqrt{2}\,\text{cis}\left( {\frac{{7\pi }}{{20}}+\frac{{8\pi \cdot 3}}{{20}}} \right)=\sqrt{2}\,\text{cis}\left( {\frac{{31\pi }}{{20}}} \right)\\k=4:\,\,\,\,\sqrt{2}\,\text{cis}\left( {\frac{{7\pi }}{{20}}+\frac{{8\pi \cdot 4}}{{20}}} \right)=\sqrt{2}\,\text{cis}\left( {\frac{{39\pi }}{{20}}} \right)\end{array}\)

 

Here’s another type of problem you might see. Note we could have solved the second part of this problem more easily with the Complex Conjugate Root Theorem, or Conjugate Zeros Theorem that we learned about here in the Imaginary (Complex) Numbers Section.

Complex Trig Root Problem Solution

Determine \(z\) and three cube roots of \(z\) if one cube root is

\(1-\sqrt{3}i\)

We can first find \(z\) by cubing  using De Moivre’s Theorem :   \(\displaystyle {{z}^{n}}={{r}^{n}}\left[ {\cos \left( {n\theta } \right)+i\sin \left( {n\theta } \right)} \right]\,\)

Convert \(1-\sqrt{3}i\) to Polar first:

\(\displaystyle r=\sqrt{{{{1}^{2}}+{{{\left( {-\sqrt{3}} \right)}}^{2}}}}=2;\,\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{{-\sqrt{3}}}{1}} \right)=\frac{{5\pi }}{3}:\,\, 2\,\text{cis}\left( {\frac{{5\pi }}{3}} \right)\)             

\(\displaystyle \begin{array}{c}{{\left[ {2\,\text{cis}\left( {\frac{{5\pi }}{3}} \right)} \right]}^{{\,3}}}={{2}^{3}}\text{cis}\left( {3\cdot \frac{{5\pi }}{3}} \right)=8\,\text{cis}\left( {5\pi } \right)=8\,\text{cis}\left( {5\pi -4\pi } \right)=8\,\text{cis}\left( \pi \right)\\=8\left( {-1+0i} \right)=-8\,\,\,\,\,\,\,\,\,\,\,\,\,z=-8\end{array}\)

Since we see that \(z=-8\), and we know that the real number cube root of 8 is 2, we can then take the conjugate of \(1-\sqrt{3}i\) to get the other root, \(1+\sqrt{3}i\) (from the Complex Conjugate Root Theorem). The roots are 2, \(1-\sqrt{3}i\), and \(1+\sqrt{3}i\).

 

But we’ll check our answer using the \(n\)th root formulas:

\(\begin{array}{c}{{\left[ {8\text{cis}\left( \pi \right)} \right]}^{{\,\frac{1}{3}}}}={{8}^{{\frac{1}{3}}}}\text{cis}\left( {\frac{{\pi +2\pi k}}{3}} \right)\\=2\,\text{cis}\left( {\frac{\pi }{3}+\frac{{2\pi k}}{3}} \right);\,\,\,\,\,k=0,1,2\end{array}\)            \(\begin{array}{l}k=0:\,\,\,\,2\,\text{cis}\left( {\frac{\pi }{3}} \right)=2\left( {\frac{1}{2}+\frac{{\sqrt{3}}}{2}i} \right)=1+\sqrt{3}i\\k=1:\,\,\,\,2\,\text{cis}\left( \pi \right)=2\left( {-1+0i} \right)=-2\\k=2:\,\,\,\,2\,\text{cis}\left( {\frac{{5\pi }}{3}} \right)=2\left( {\frac{1}{2}-\frac{{\sqrt{3}}}{2}i} \right)=1-\sqrt{3}i\end{array}\)

 

How to Check with Graphing Calculator

When getting roots of complex numbers in the calculator, you’ll only get one root, so this won’t be too helpful. But you can check all the roots you get by putting them in the calculator and taking the reverse power.

For example, if we try to get the roots for \(\sqrt[3]{{-27}}\), we only get –3. But to check our answers that we got above, we can cube them to see if we get –27:

Roots in Calculator

To check to see if 216° and 288° are fifth roots of unity (1), as other examples, we can convert them to radians and take the 5th root:

Roots in Calculator 2

(Note that the coefficient of \(i\) is a very tiny number and should be in fact “0”.)

Hint: You can always check the complex roots by entering, for example, cube root of -27 in Wolfram Alpha.

Understand these problems, and practice, practice, practice!


Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

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On to Introduction to Calculus and Study Guides  – you’re ready!

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