# 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= $\pi /6$,$5 \pi /6$,$9 \pi /6$ ? – Jasser Oct 23 '14 at 9:59
• Notice that $\frac{\pi}{6}+\frac{2\pi}{3}=\frac{5\pi}{6}$. Except that, your solution is correct. – Galc127 Oct 23 '14 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 $k \in [0,n-1]$. – Galc127 Oct 23 '14 at 10:25

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.

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) – Galc127 Oct 23 '14 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 '14 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. – UserX Nov 1 '14 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 '14 at 11:33
• @Did $z^3-i=z^3+i^3$ – UserX Nov 1 '14 at 11:40

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 '14 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. – Martigan Oct 23 '14 at 10:06