Question

How do you convert $-8+8i$ to polar form?

Answer 1

Hint: We will try to convert to polar form by determining the modulus form and the angle of the complex number $-8+8i$. The polar form $z=x+iy$ is $z=r\left( \cos \alpha +i\sin \alpha \right)$. We need to take the angle of the complex number according to its position on the plane.

Complete step-by-step answer:
We will follow the process of converting regular form of complex numbers to their polar form.
Let’s assume the general form of complex number is $z=x+iy$.
Here $i$ represents the complex number and $x$ and $y$ are real constants.
Then we find the modulus value for the complex number as \[r=\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\].
Also, to find the angle $\alpha $ for the we can say that $x=r\cos \alpha ,y=r\sin \alpha $.
The simplest form for the angle $\alpha $ is $\tan \alpha =\dfrac{y}{x}$.
Therefore, the polar form representation of the complex number $z=x+iy$ is
$z=r\left( \cos \alpha +i\sin \alpha \right)$.
Now for our given complex number $-8+8i$, we equate it with $z=x+iy$.
We try to find the polar form of the $-8+8i$.
Equating values, we get $x=-8,y=8$.
The modulus value for the complex number will be \[r=\sqrt{{{\left( -8 \right)}^{2}}+{{8}^{2}}}=8\sqrt{2}\].
Now we find the angle value which is $\alpha $.
The representation of the $-8+8i$ gives the point at the second quadrant.
This means the angle will be in the interval of $\dfrac{\pi }{2}\le \alpha \le \pi $
We get $\tan \alpha =\dfrac{y}{x}=\dfrac{8}{-8}=-1$. We need to find the exact solution for the inverse where the angle is in second quadrant.
So, $\tan \alpha =-1=\tan \left( \dfrac{\pi }{2}+\dfrac{\pi }{4} \right)$. The angle is $\left( \dfrac{3\pi }{4} \right)$.
Now we form the polar form and get \[-8+8i=8\sqrt{2}\left[ \cos \left( \dfrac{3\pi }{4} \right)+i\sin \left( \dfrac{3\pi }{4} \right) \right]\].

Note: The usual angle value for $\tan \alpha =-1$ would have been $\left( -\dfrac{\pi }{4} \right)$. But the position of the complex number is in the angle interval of $\dfrac{\pi }{2}\le \alpha \le \pi $. That’s why we had to place the angle in that position to get the value of $\tan \alpha =-1$.