# Solve complex equation $z^3 = i$

I have this $z^3 = i$ complex equation to solve.

I begin with rewriting the complex equation to $a+bi$ format.

1 $z^3 = i = 0 + i$

2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$

3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.

4 The complex equation can now be rewriten $w^3=r^3(cos3v+i\sin3v)$, $w^3 = 1^3(\cos \frac {\pi}{2} 3 +i \sin \frac {\pi}{2} 3)$ or $w^3 = e^{i \frac {\pi}{2} 3}$.

5 Calculate the angle $3 \theta = \frac {\pi}{2} + 2 \pi k$ where $k = 0, 1, 2$

6 $k = 0$, $3 \theta = \frac {\pi}{2} + 2 \pi 0 = \frac {\pi}{6}$

7 $k = 1$, $3 \theta = \frac {\pi}{2} + 2 \pi 1 = \frac {\pi}{6} + \frac {2 \pi}{3} = \frac {5 \pi}{6}$

8 $k = 2$, $3 \theta = \frac {\pi}{2} + 2 \pi 2 = \frac {\pi}{6} + \frac {4 \pi}{3} = \frac {9 \pi}{6}$

So the angles are $\frac {\pi}{6}, \frac {3 \pi}{6}, \frac {9 \pi}{6}$ but that is no the correct answer. The angle of the complex equation should be $-\frac {\pi}{2}$ where I calculated it to $\frac {\pi}{2}$. I'm I wrong or is there a mistake in the book I'm using?

Thanks!

• Isn't k= $\pi /6$,$5 \pi /6$,$9 \pi /6$ ? Oct 23, 2014 at 9:59
• Notice that $\frac{\pi}{6}+\frac{2\pi}{3}=\frac{5\pi}{6}$. Except that, your solution is correct. Oct 23, 2014 at 9:59
• BTW, if you get an equation of the form $\displaystyle z^n=r e^{i\theta}$ you can always use the formula $\displaystyle z_k=\sqrt[n]{r} \large e^{\frac{i(\theta+2\pi k)}{n}}$ where $k$ is an integer such that $k \in [0,n-1]$. Oct 23, 2014 at 10:25

Way easier way;

$$z^3=i \\ \iff z^3-i=0 \\ \stackrel{-i=i^3}{\iff}z^3+i^3=0 \\ \iff (z+i)(z^2-iz-1) = 0 \\ \iff z_1=-i,\; z_2=\frac12 (i-\sqrt 3), \; z_3=\frac12 (i+\sqrt 3)$$

• Nice answer (+1) Oct 23, 2014 at 10:29
• This 1. requires to guess that $-i$ is a root and 2. does not answer the only question I see in the OP.
– Did
Nov 1, 2014 at 11:19
• @Did 1. You can see it as an identity(which is how I thought of it too). 2. But it CAN help OP in a way. If you disagree you can always flag/downvote the answer. Nov 1, 2014 at 11:27
• "You can see it as an identity(which is how I thought of it too)." Dunno what you mean. "If you disagree you can always flag/downvote the answer" Sorry?
– Did
Nov 1, 2014 at 11:33
• @Did $z^3-i=z^3+i^3$ Nov 1, 2014 at 11:40

Step $4$ is where your mistake happens. Your original equation is

$$z^3=i$$

Then you rewrite $i=1\cdot(\cos\frac\pi2 + i\sin\frac\pi2)$ and rewrite $z = r(\cos v + i\sin v)$, meaning that $$z^3=i$$ will change into $$r^3(\cos3v + i\sin 3v) = 1\cdot(\cos\frac\pi2 + i\sin\frac\pi2)$$

What you made was you also took the third power of $i$, which was wrong.

Use polar coordinates.

$z^{3}=i=e^{i(\frac{\pi}{2}+2k\pi)}$, $k\in \mathbb{Z}$

And from here it is much simpler

EDIT: what I mean by much simpler...

The OP kind of used polar form, but not really. You should stock with polar form until the very end. The OP introduces $\theta$, $cosinus$ and $sinus$ functions etc. You mix everything up and forget that $(\frac{\pi}{2}+2\pi)$, divided by $3$, gives $\frac{5\pi}{6}$...

• OP did do that. He is asking what he did wrong.
– 5xum
Oct 23, 2014 at 10:03
• @5xum Well, he did and did not... He did it in a very complex way, denoting that his mind was not clear, and leading to error in its calculus... If you take it all the way in polar form, you are limiting the possible errors. Oct 23, 2014 at 10:06