solve complex equation

• Mar 31st 2018, 10:22 AM
s_ingram
solve complex equation
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?
• Mar 31st 2018, 10:41 AM
Idea
Re: solve complex equation
$\left|z|^4=4\right.$

$|z|=\sqrt{2}$

$z=\sqrt{2}(\cos t + i \sin t)$
• Mar 31st 2018, 12:24 PM
HallsofIvy
Re: solve complex equation
Your error was in writing z as cos(t)+ i sin(t). That is true only for z having absolute value 1. But, as Idea points out, in order that z satisfy $\displaystyle z^4= 4$, z must have absolute value $\displaystyle \sqrt[4]{4}= \sqrt{2}$.

You could write z as rcos(t)+ ri sin(t) for real r and t.