Question
Convert the complex number in the polar form: $-3$

Answer

Here $z = -3 = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\;\cos \theta = - 3$ and $r\;\sin \theta = 0$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 9 + 0$$ \Rightarrow {r^2} = 9 \Rightarrow r = 3$
$\therefore 3\cos \theta = - 3$ and $3\sin \theta = 0$
$ \Rightarrow \cos \theta = - 1$ and $\sin \theta = 0$
Since $\sin \theta $ is positive and $\cos \theta $ is negative
$\therefore \theta $ lies in second quadrant
$\therefore\theta=(\pi-0)\;=\;\mathrm\pi$
Hence polar form of z is $3(\cos \pi + i\sin \pi )$

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