Solve the equation z^4 + 81 = 0
$\displaystyle z^4+81=0$Originally Posted by norivea
$\displaystyle z^4=-81$
Taking the square root of both sides:
$\displaystyle z^2 = \pm 9i $
Again taking the square root of both sides:
$\displaystyle z = \pm \sqrt{\pm 9i}$ = $\displaystyle \pm 3 \sqrt{\pm i}$
So what is $\displaystyle \sqrt{\pm i}$?
To answer this, let's write "i" a bit differently using the exponential form for complex numbers:
$\displaystyle i = cos(\pi/2)+i \, sin(\pi/2) = e^{i\pi/2}$
Thus
$\displaystyle \sqrt{i}=(i)^{1/2} = \left( e^{i\pi/2} \right ) ^{1/2}$ = $\displaystyle \left( e^{i\pi/2*1/2} \right ) = e^{i\pi/4}$ = $\displaystyle cos(\pi/4)+ i \, sin(\pi/4) = \frac{\sqrt2}{2} + i \frac{\sqrt2}{2}$ = $\displaystyle \frac{\sqrt2}{2}(1+i)$
Similarly:
$\displaystyle \sqrt{-i}=(-i)^{1/2} = \left( e^{-i\pi/2} \right ) ^{1/2}$ = ... = $\displaystyle \frac{\sqrt2}{2}(1-i)$
So finally the solution to $\displaystyle z^4+81=0$ is
$\displaystyle z =\pm 3 \sqrt{\pm i}$ =$\displaystyle \pm 3 \frac{\sqrt2}{2}(1+i)$ and $\displaystyle \pm 3 \frac{\sqrt2}{2}(1-i)$
-Dan
Hello, norivea!
Solve the equation: $\displaystyle z^4 + 81 \:= \:0$
We have: .$\displaystyle z^4\:=\:-81$
The polar form of $\displaystyle -81$ is: .$\displaystyle 81(\cos\pi + i\sin\pi)$
Then: .$\displaystyle z\;=\;[81(\cos\pi + i\sin\pi)]^{\frac{1}{4}}$
Hence: .$\displaystyle z\;=\;3\bigg[\cos\left(\frac{\pi}{4} + \frac{\pi}{2}n\right) + i\sin\left(\frac{\pi}{4} + \frac{\pi}{2}n\right)\bigg]$ for $\displaystyle n = 0,1,2,3$
Therefore: .$\displaystyle z\;=\;\left\{\frac{3}{\sqrt{2}}(1 + i),\;\frac{3}{\sqrt{2}}(1 - i),\;\frac{3}{\sqrt{2}}(-1 + i),\;\frac{3}{\sqrt{2}}(-1 - i)\right \}$
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Alternate method
We have: .$\displaystyle z^4\:=\:-81\quad\Rightarrow\quad z^2 \:=\:\pm 9i\quad\Rightarrow\quad z\:=\:\pm\sqrt{\pm9}\sqrt{i}$
. . Then: .$\displaystyle z\:=\:\pm3\sqrt{i}$ and $\displaystyle \pm3i\sqrt{i}$
Now, if you happen to know that: .$\displaystyle \sqrt{i}\:=\:\frac{1 + i}{\sqrt{2}}$
. . then we have: .$\displaystyle z\:=\:\pm 3\left(\frac{1 + i}{\sqrt{2}}\right)$ and $\displaystyle \pm 3i\left(\frac{1 + i}{\sqrt{2}}\right)$
which gives us the same fourth roots.
$\displaystyle \begin{aligned} x^4+3^4 & = (x^2+3^2)^2-2\cdot3^2x^2 = (x^2+3^2)^2-(3\sqrt{2}x)^2 = (x^2-3\sqrt{2}x+3^2)(x^2+3\sqrt{2}x+3^2). \end{aligned}$
From here I trust you know how to solve quadratic equations (use the quadratic formula/complete the square).