Question

Z=4-5i using polar form find z^2

Answer 1

\[z=a+bi\]
In polar form, it would be in a form like this:
\[z=r(\cos \theta + isin \theta)\]\[r=\sqrt{a^2 + b^2}\]\[\theta=\tan^{-1} (b/a)\]
so for this case: \[r=\sqrt{4^2 + (-5)^2}=\sqrt{41}\]\[\theta=\tan^{-1} (-5/4) = -0.896\]
so to find z^2 is quite straight forward:
\[z^2 = r^2 ( \cos2\theta + isin2\theta)\]
And remember that the range of theta should be :
\[-\pi<\theta<\pi\]