Questions:
a) Express $-4i$ in polar form
b) Hence solve the equation $z^2 =-4i$
I've attempted the question and was only able to find $z=4$ using the form $r(\cos x+i\sin x)=z$.
Could someone please show me or explain how to do step by step?
Thankyou
Best Answer
Are you familiar with Argand diagrams, if so I recommend making a quick sketch.
We use the relation $$z=(x+iy)=r\left(\cos(\theta)+i\sin (\theta)\right)$$ $|4i|=4$. This is the magnitude of the complex vector
$\arg(z)=\tan^{-1}\left(\cfrac{y}{x}\right)$. In your case $x=0$ and $y=-4$.
$\arg(-4i)=\tan^{-1}\left(\cfrac{-4}{0}\right)=\tan^{-1}(-\infty)=-\cfrac{\pi}{2}=\theta$. This is the angle the complex vector makes with the real-axis on an Argand diagram. Obviously taking the inverse tan of infinity is not the best way to do it as $\infty$ is undefined so making a sketch on an Argand Diagram makes more sense.
So $-4i=4(\cos(-\frac{\pi}{2})+i\sin (-\frac{\pi}{2}))=-4i$
Now that we know the argument and magnitude we can move on to solve part b)
First rewrite $z^2 =-4i$ as $z^2 +4i =0$ and this time $\arg(z^2)=2k\pi-\cfrac{\pi}{2}$ as any integer $k$ of $2\pi$ yields the same result. Then one way of finding the solutions is to use $$z^n=(x+iy)^n=r^ne^{in\theta}=r\left(\cos(n\theta)+i\sin (n\theta)\right)$$ which is known as De Moivres Theorem. But if you have not come across this it might be best to look it up. So we have $$z=r^{\frac{1}{n}}\left(\cos(\theta)+i\sin (\theta)\right)^\frac{1}{n}= 4^{\frac{1}{2}}\left(\cos(-\cfrac{\pi}{2}+2k\pi)+i\sin (-\cfrac{\pi}{2}+2k\pi)\right)^\frac{1}{2}= 2\left(\cos\left(\cfrac{-\frac{\pi}{2}+2k\pi}{2}\right)+i\sin \left(\cfrac{-\frac{\pi}{2}+2k\pi}{2}\right)\right)$$ and this is valid for $k=0$ and $k=1$ as there are two roots to the equation. So now you simply substitute these values of $k$ into the formula to get the two solutions. After you've done that you should get answers of $z=\sqrt2-\sqrt2i$ and $z=\sqrt2i-\sqrt2$. I have omitted the simplification steps that lead to these results for $z$ as it takes up too much space (and time). But I can show you if you like.
It's helpful to notice that $z^2=(\sqrt2-\sqrt2i)(\sqrt2i-\sqrt2)= 2i -2 -2i^2+2i=2i -2+2+2i=4i$ which is precisely what we would expect.