Solve the equation z^4 + 81 = 0

**norivea** · June 14th 2006, 03:21 AM

**topsquark** · June 14th 2006, 03:43 AM

Originally Posted by norivea

Solve the equation z^4 + 81 = 0

$z^4+81=0$

$z^4=-81$

Taking the square root of both sides:
$z^2 = \pm 9i$

Again taking the square root of both sides:
$z = \pm \sqrt{\pm 9i}$ = $\pm 3 \sqrt{\pm i}$

So what is $\sqrt{\pm i}$ ?

To answer this, let's write "i" a bit differently using the exponential form for complex numbers:
$i = cos(\pi/2)+i \, sin(\pi/2) = e^{i\pi/2}$

Thus
$\sqrt{i}=(i)^{1/2} = \left( e^{i\pi/2} \right ) ^{1/2}$ = $\left( e^{i\pi/2*1/2} \right ) = e^{i\pi/4}$ = $cos(\pi/4)+ i \, sin(\pi/4) = \frac{\sqrt2}{2} + i \frac{\sqrt2}{2}$ = $\frac{\sqrt2}{2}(1+i)$

Similarly:
$\sqrt{-i}=(-i)^{1/2} = \left( e^{-i\pi/2} \right ) ^{1/2}$ = ... = $\frac{\sqrt2}{2}(1-i)$

So finally the solution to $z^4+81=0$ is

$z =\pm 3 \sqrt{\pm i}$ = $\pm 3 \frac{\sqrt2}{2}(1+i)$ and $\pm 3 \frac{\sqrt2}{2}(1-i)$

-Dan

**Soroban** · June 14th 2006, 04:23 AM

Hello, norivea!

Solve the equation: $z^4 + 81 \:= \:0$

We have: . $z^4\:=\:-81$

The polar form of $-81$ is: . $81(\cos\pi + i\sin\pi)$

Then: . $z\;=\;[81(\cos\pi + i\sin\pi)]^{\frac{1}{4}}$

Hence: . $z\;=\;3\bigg[\cos\left(\frac{\pi}{4} + \frac{\pi}{2}n\right) + i\sin\left(\frac{\pi}{4} + \frac{\pi}{2}n\right)\bigg]$ for $n = 0,1,2,3$

Therefore: . $z\;=\;\left\{\frac{3}{\sqrt{2}}(1 + i),\;\frac{3}{\sqrt{2}}(1 - i),\;\frac{3}{\sqrt{2}}(-1 + i),\;\frac{3}{\sqrt{2}}(-1 - i)\right \}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Alternate method

We have: . $z^4\:=\:-81\quad\Rightarrow\quad z^2 \:=\:\pm 9i\quad\Rightarrow\quad z\:=\:\pm\sqrt{\pm9}\sqrt{i}$

. . Then: . $z\:=\:\pm3\sqrt{i}$ and $\pm3i\sqrt{i}$

Now, if you happen to know that: . $\sqrt{i}\:=\:\frac{1 + i}{\sqrt{2}}$

. . then we have: . $z\:=\:\pm 3\left(\frac{1 + i}{\sqrt{2}}\right)$ and $\pm 3i\left(\frac{1 + i}{\sqrt{2}}\right)$

which gives us the same fourth roots.

**TheSaviour** · January 15th 2013, 10:08 AM

$\begin{aligned} x^4+3^4 & = (x^2+3^2)^2-2\cdot3^2x^2 = (x^2+3^2)^2-(3\sqrt{2}x)^2 = (x^2-3\sqrt{2}x+3^2)(x^2+3\sqrt{2}x+3^2). \end{aligned}$

From here I trust you know how to solve quadratic equations (use the quadratic formula/complete the square).

**topsquark** · January 15th 2013, 12:11 PM

Originally Posted by TheSaviour

$\begin{aligned} x^4+3^4 & = (x^2+3^2)^2-2\cdot3^2x^2 = (x^2+3^2)-(3\sqrt{2}x)^2 = (x^2-3\sqrt{2}x+3^2)(x^2+3\sqrt{2}x+3^2). \end{aligned}$

From here I trust you know how to solve quadratic equations (use the quadratic formula/complete the square).

Nice idea. One slight typo: The first term past the second = sign should be squared.

Ummmm...No problem with you helping out, but you are aware that this thread is over 6 years old?

-Dan

**TheSaviour** · January 15th 2013, 12:16 PM

Originally Posted by topsquark

Nice idea. One slight typo: The first term past the second = sign should be squared.

Ummmm...No problem with you helping out, but you are aware that this thread is over 6 years old?

-Dan

LOOOOOOL! I honestly had no idea. How on earth did it show up on the main/home pages?

EDIT: Oh, I think I know why. I was looking Who's Online and someone was reading this! Trap!

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