# Find $z$ when $z^4=-i$?

Consider $z^4=-i$, find $z$.

I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$

$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$

$|z^4|=\sqrt{(-1)^2}=1$

$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$

$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\frac{\pi}{8}$

Since $z^n=r^n\cdot e^{in\theta}$, $z^4$ can now be rewritten as $z^4=e^{i\cdot4\cdot\frac{\pi}{8}} \iff z=e^{i\frac{\pi}{8}}$ However, my answer file says this is wrong. Can anyone give me a hint on how to find $z$?

• $\arg(-i)=3\pi/2$, not $\pi/2$
– Paul
Sep 4 '15 at 11:44

$$\left(e^{i\frac\pi8}\right)^4 = e^{i\frac\pi2} = i$$

so you obviously made some error.

The error stemms from the line

$$\cos(4\theta) = 0\iff \arccos(0) = 4\theta$$

which is not true for all values of $\theta$. For example, $\cos(4\cdot \frac{3\pi}{8}) = 0$, even though $\arccos(0)\neq 4\cdot \frac{3\pi}{8}$.

In fact, there are four distinct solutions of $\theta$ for the equation

$$\cos(4\theta) + i\sin(4\theta)= 0 + (-1)\cdot i$$

One of these solutions can be found through using the $\arcsin$ and $\arccos$ functions, the others can then be found by noticing that you can replace $\theta$ with $\theta + \frac{2k\pi}{4}$ and still satisfy the equation, as long as $k\in\mathbb Z$.

the last step, you need $$-i = \cos(3\pi/2) + \sin (3\pi/2) = \cos (4 \theta) + i \sin(4 \theta)$$ which will give you $$4\theta = 3\pi/2, 3\pi/2 + 2\pi, 3\pi/2 + 4\pi, 3\pi/2 + 6\pi$$ now you can solve for $\theta.$

Remember, that $-i = e^{i\frac{3\pi}{2}}$, then

$$z^4 = -i = e^{i\frac{3\pi}{2}}$$

The n$^{\text{th}}$ root of a complex number is

$$z^n = re^{i\varphi} \implies z_k = r^{1/n}e^{i\left(\frac{\varphi}{n} + \frac{2\pi k}{n} \right)} \qquad \text{where} \quad k = 0, \dots, n-1$$

$$z_0 = e^{i\left(\frac{3\pi}{8} + 0 \right)} = (r, \varphi_0) = \left(1, \frac{3 \pi}{8}\right)$$ $$z_1 = e^{i\left(\frac{3\pi}{8} + \frac{2\pi}{4} \right)} = (r, \varphi_1) = \left(1, \frac{7 \pi}{8}\right)$$ $$z_2 = e^{i\left(\frac{3\pi}{8} + \frac{4\pi}{4} \right)} = (r, \varphi_2) = \left(1, \frac{11 \pi}{8}\right)$$ $$z_3 = e^{i\left(\frac{3\pi}{8} + \frac{6\pi}{4} \right)} = (r, \varphi_3) = \left(1, \frac{15 \pi}{8}\right)$$
We have $z^4=e^{3\pi i/2+2k\pi}$.
So $z=(e^{3\pi i/2+2k\pi})^{1/4}=e^{3\pi i/8+k\pi/2}$
As $e^{i\theta}=e^{i(\theta+2\pi)}$, the $4$ values of $z$ are $e^{3\pi i/8}, e^{7\pi i/8}, e^{11\pi i/8}, e^{15\pi i/8}$.