Question

How do you solve this equation over the complex numbers?

z^4+36=0

[SOLUTIN IN ALGEBRIC FORM]

Answer 1

subtract 36 from both sides, and you get \[z^4=-36\]take the square root of both sides.
\[z^2=\sqrt{36}~~~~~->~~~~~z^2=\sqrt{6} \times \sqrt{6} \times i~~~~~->~~~~~z^2=6i \]
take the square root of both sides again.
\[z=\sqrt{ 6i }\]

Answer 2

Do you know what "i" is ?

Answer 3

Yes, I do know. There are 4 different roots.. I have managed to get z^4=6i.
Don't know what to do next.. need to use De Moiver?

Answer 4

I forgot + -6i, and to then square both sides, sorry.

Answer 5

I don't know what De movier is, sorry again

Answer 6

\[
z= r e^{i t}\\

r^4 e^{4 i t} = 36 e^{ i \pi}\\\
r=\sqrt{6}\\
4 t = \pi + 2 k \pi\\
t= \frac {\pi}4+ k \frac{\pi}2\\
t=\frac \pi 4\\
t=\frac \pi 4 + \frac \pi 2=3 \frac \pi 4\\
t=\frac \pi 4 + 2 \frac \pi 2=5 \frac \pi 4\\
t=\frac \pi 4 + 3 \frac \pi 2=7 \frac \pi 4\\

\]
You replace each t and r by their values and you obtain the three roots

Answer 7

four roots