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The equation z^5 = i has 5 solutions.

The unique solution in the third quadrant is re^(i theta), where r > 0 and 0 =< theta < 2pi. What is the ordered pair (r, theta).

Apr 22, 2019
edited by Guest  Apr 22, 2019

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Is the unique solution supposed to be $$re^{i\theta}$$ or $$re^{\frac{i}{\theta}}$$?

Apr 22, 2019
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I'm in precalculus and we just finished learning this stuff so sorry if i'm wrong.

So if your wondering where $$re^{i\theta}$$ comes from then let me explain! (its kinda a long explanation) This is part of the larger answer so even if you don't you should still read it.

The rectangular form of a complex number is $$a+bi \hspace{0.1cm}\textrm{such that} \hspace{0.1cm} a,b \in \mathbb{R}$$ This means that a and b are real numbers that can be manipulated as normal. a and b in the complex plane are what you might typically associate with x and y for two real numbers in the real plane forming the ordered pair (x,y).

Just as ordered pairs in the real plane can be expressed in both rectangular and polar forms as per:

Rectangular to Polar

$$(x,y) \rightarrow (\sqrt{x^2+y^2}, tan^{-1} (\frac{y}{x}))$$

Remember that when x is negative, to add π to your angle measure when using radians or 180 degrees when using degrees.

Polar to Rectangular

$$(r, \theta) \rightarrow (r \hspace{0.1cm}cos \hspace{0.1cm}\theta, r \hspace{0.1cm} sin \hspace{0.1cm}\theta)$$

Complex numbers in the complex plane can be expressed in rectangular and polar forms too! They are expressed as follows:

Rectangular to Polar

$$a+bi \rightarrow \sqrt{a^2+b^2} (cos (tan^{-1}(\frac{b}{a}))+i\sqrt{a^2+b^2}(sin(tan^{-1}(\frac{b}{a}))$$

Which can be rewritten factored as $$\sqrt{a^2+b^2}(cos(tan^{-1}(\frac{b}{a}))+i\hspace{0.05cm}sin(tan^{-1}(\frac{b}{a})))$$

With substitution it can also be written as $$r(cos\hspace{0.1cm} \theta + i \hspace{0.05cm}sin \hspace{0.1cm}\theta)$$ where $$r = \sqrt{a^2+b^2} \hspace{0.1cm} \textrm{and} \hspace{0.1cm} \theta = tan^{-1}(\frac{b}{a})$$

The values of r and theta are derived using geometry and are the same as with real numbers (mathematically speaking, this means not only do the axioms of standard euclidean geometry (and so also the theorems of geometry) hold up in the real plane but they also hold up in the complex plane)

$$\sqrt{a^2+b^2}$$ is found by using the distance formula $$D(P_1,P_2)=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$$ where $$P_1 \hspace{0.1cm} \textrm{is the origin} \hspace{0.1cm} (x_1=0,y_1=0)$$ and $$P_2 = (x_2=a,y_2=b)$$ where these ordered pairs represent points in rectangular form.

$$tan^{-1}(\frac{b}{a})$$ is found by plotting Point 2 and using right triangle trigonometry to find that the tangent of theta is the ratio of leg b to leg a. Solving for theta gives the formula. Because of the domain restriction on inverse tangent, it is important to remember to add π or 180 when a is negative.

Polar to Rectangular

Just simplify $$r cos \theta + i \hspace{0.1cm} rsin \theta$$ in your calculator to get $$a+bi$$

HAVING ALL OF THAT SAID $$re^{i\theta}$$ is just a fancy way of writing a solution to the equation your trying to solve and we get this way of writing one of the solutions with euler's formula.

$$e^{i\theta}=cos (\theta) + i \hspace{0.1cm}sin (\theta)$$

The polar form of a complex number is $$r cos(\theta)+i \hspace{0.1cm} r sin(\theta) = r(cos(\theta)+i\hspace{0.1cm}sin(\theta))$$ and so you get $$re^{i\theta}=r(cos(\theta)+i\hspace{0.1cm}sin(\theta))$$

So the equation your trying to solve is $$z^5=i$$

They ask for $$re^{i\theta}$$ (or the solution) that is specifically in the third quadrant. How do you know which one that is? We know right now there are five solutions from the fundamental theorem of algebra by the way.

If you visit this link of the graph I made at desmos https://www.desmos.com/calculator/xvxq8e8mmk You will see five points on a unit circle (circle with radius of 1 centered at the origin). Those five points are the solutions to your equation. You will notice there is only 1 point in the third quadrant, that is the unique solution you are looking for. You must imagane that the graph is over the complex plain and not the real plane. (the same geometric rules apply as we discussed earlier so they for all intents and purposes are essentially the same).

So how do you solve the equation $$z^5=i$$? to start it might help to take the fifth root of both sides.

$$\sqrt{z^5}=\sqrt{i}$$

$$z=\sqrt{i}$$

So z is a complex number and assuming it is in polar form... that is $$r(cos(\theta)+i\hspace{0.1cm}sin(\theta))$$ then your solution is given by the following formula:

$$z_k=\sqrt[n]{r}\bigg[cos\bigg(\frac{\theta}{n}+\frac{2\pi k}{n}\bigg)+i \hspace{0.1cm} sin\bigg(\frac{\theta}{n}+\frac{2\pi k}{n}\bigg)\bigg]$$

Where n is the nth root of the complex number and k is one of the n solutions. The trickest part of this formula is the k. Instead of the first solution having a k of 1 it has a k of 0 and instead of the second solution having a k of 2 it has a k of 1. In general the mth solution has a k of $$m-1$$

So for fifth root of a complex number, the solutions in order would have k values of 0, 1, 2, 3, 4

In general solutions to roots of complex numbers have complex values of 0, 1, 2, ... , $$n-1$$

So the first step in using this formula would be converting our complex number to polar form. $$0+i = 1(cos \frac{π}{2} + i\hspace{0.1cm} sin \frac{π}{2})$$ You might notice the inverse tangent formula does not work. In this case because the point lies on an axis, the angle will be quadrantal and so the inverse tangent formula either gives wrong answers or no answer at all. Just graph the point in the complex plane and you can clearly see the angle is 90 degrees or half of π.

The second step is using the formula to find each solution. Note I write the solutions in degrees to save time not having to convert to radians, also the graph I shared with you will be easier to visuallize.

$$z_0=\sqrt{1}\bigg[cos\bigg(\frac{90}{5}+\frac{360*0}{5}\bigg)+i \hspace{0.1cm} sin\bigg(\frac{90}{5}+\frac{360*0}{5}\bigg)\bigg]$$$$= cos(18)+i \hspace{0.1cm} sin(18)$$

$$z_1=\sqrt{1}\bigg[cos\bigg(\frac{90}{5}+\frac{360*1}{5}\bigg)+i \hspace{0.1cm} sin\bigg(\frac{90}{5}+\frac{360*1}{5}\bigg)\bigg]$$$$= cos(90)+i \hspace{0.1cm}sin(90)$$

$$z_2=\sqrt{1}\bigg[cos\bigg(\frac{90}{5}+\frac{360*2}{5}\bigg)+i \hspace{0.1cm} sin\bigg(\frac{90}{5}+\frac{360*2}{5}\bigg)\bigg]$$$$= cos(162)+i \hspace{0.1cm} sin(162)$$

$$z_3=\sqrt{1}\bigg[cos\bigg(\frac{90}{5}+\frac{360*3}{5}\bigg)+i \hspace{0.1cm} sin\bigg(\frac{90}{5}+\frac{360*3}{5}\bigg)\bigg]$$$$=cos(234)+i \hspace{0.1cm} sin(234)$$

$$z_4=\sqrt{1}\bigg[cos\bigg(\frac{90}{5}+\frac{360*4}{5}\bigg)+i \hspace{0.1cm} sin\bigg(\frac{90}{5}+\frac{360*4}{5}\bigg)\bigg]$$$$= cos(306)+i \hspace{0.1cm} sin(306)$$

So I just thought I might point out something interesting but it is not relivant to your problem.

How often do you find a solution to the equation $$\sqrt{x}=x$$?

We just did! Even if it is complex that is still pretty cool. $$cos(90)+i \hspace{0.1cm} sin(90) = i$$ & so $$\sqrt{i}=i$$ by the formula we just used. You might have learned previously that $$i^5=i$$ as $$i^4 =1$$

i must be a fifth root for i to the fifth power to be i as

$$\sqrt{x}\sqrt{x}\sqrt{x}\sqrt{x}\sqrt{x}=x$$

and so this actually makes sense! YAY MATH!

Moving on back to your problem. How do we know which of these five solutions is in quadrant three? Just look at your theta. 234 degrees is the only theta between 180 degrees and 270 degrees and so is the only solution in the third quadrant. I created my graph by writing the complex solutions as real ordered pairs $$(r cos (\theta), r sin (\theta))$$

The r is root 5 of 1 or just 1 and the theta angle measure is 234 degrees or $$\frac{13\pi}{10}$$ and so both the r and theta meet your restrictions of being a positive r and a theta between 0 and 2 pi.

As for the final part of your question (r, theta)?

It is important to relize that the ordered pair in question and the complex number solution we arrived at ARE NOT EQUIVALENT. One is in the real plane and the other is in the complex plane. Should you want to convert, the answer is in fact $$(1, \frac{13\pi}{10})$$

If you have any questions feel free to ask!

Apr 22, 2019