Hi folks,

I am trying to solve the complex equation $z^4 + 4 = 0$

I did it by setting $z = a + i b$ and the results came out. 1 + i, 1 - i, -1 + i, -1 - i.

I then tried to do the same thing putting z in polar form and have no such luck.

$(\cos t + i \sin t)^4 + 4 = 0$ gives

$ \cos 4t + 4 + i \sin 4t = 0$ using de Moivre's theorem.

so $ \cos 4t + 4 = 0 $ ........ real

and $ \sin 4t = 0 $ ........ imaginary

the imaginary part gives $4t = n \pi$ i.e. $t = \dfrac{n\pi}{4} $

when $n = 0$, $z = 1$

when $n = 1$, $z = \dfrac{1}{\sqrt{2}} + i\dfrac{1}{\sqrt{2}}$

when $n = 2$, $z = i$

the real part gives

$\cos 4t + 4 = 0$

it can be shown that $\cos 4t = 8\cos^4 t- 8\cos^2 t + 1$

$8\cos ^4 t - 8 \cos^2 t + 5 = 0$ let $x = \cos^2 t$

$8x^2 - 8x + 5 = 0$

$x = \dfrac{4 \pm i \sqrt{14}}{8} $ and this looks nothing like the correct answer.

I can't see any mistakes in my work, but there is no way this is getting to the right answer.

Can anyone see what I am doing wrong?