How to take the root of a complex number when the right side of the equation is zero?

Question

so I know that when wanting to take the root of the complex number in a case like this

$$z^4=1+\sqrt3i$$

it's really just computing the polar coordinates and finding the solutions for $z_{0}, z_{1},...,z_{3}$ with De Moivre's formula etc.

but when we have something like this instead...

$$z^7+7i=0$$

how am I supposed to compute the $\Re$ and $\Im$ parts of this? With the first exercise I can easily come up with the polar form, but the second one is just horrendous. I don't see where that number has its real and imaginary parts which makes computing $r$ and especially the angle $\varphi$ a mystery.

I would really appreciate any help ;-; thank you so much!!

Answer 1

$\theta = (-\pi/2) \implies -i = e^{i\theta} \implies z^7 = -7i = 7e^{i\theta}$.

Therefore, one of the roots of the equation is
$z_r = (7)^{(1/7)}e^{i(\theta/7)}.$

Having found one of the roots, you find all 7 roots, by computing
$w_k = z_r \times z_k : k \in \{0,1,2,3,4,5,6\}$
where $z_k$ is the $k$-th root of $z^7 = 1 \implies $
$z_k = e^{(i2k\pi/7)}.$

Edit
To elaborate:

• $e^{i\theta}$ is shorthand for $[\cos(\theta) + i\sin(\theta)].$
• Although (for example) $\cos(-\pi/14)$ is probably expressible by radicals, this is almost certainly not the point of the problem. Therefore, expressing the roots by their polar coordinates should be fine.

Answer 2

You can write it as $$z^7 = -7 \iota$$and then take modulus to get $r^7 = 7$

Here $\iota = \sqrt{-1}$

Answer 3

HINT

Simply rearrange into the form $$z^7=-7i$$ We can let $z=r(\cos(\theta+2k\pi)+i\sin(\theta+2k\pi))$ and proceed like usual. If you need any more help please don't hesitate to ask.