Question

How do you find the principal argument of $z=3i$ $?$

Answer 1

Finding the principal argument:

Step 1: find the principal argument of complex number

$z=x+iy$ which can be written as $z=r\cos \left(\theta \right)+ir\sin \left(\theta \right)$

where $x=r\cos \left(\theta \right)$ and $y=r\sin \left(\theta \right)$

Step 2: Substituting the values

$z=3i$

$0=r\cos \left(\theta \right)$ and $3=r\sin \left(\theta \right)$

By squaring and adding the above values we get,

$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=9$

$$\begin{gathered} r=\sqrt{9} \\ =3 \end{gathered}$$

$$\begin{gathered} \cos \left(\theta \right)=0,\sin \left(\theta \right)=1 \\ \theta =\frac{\pi }{2} \end{gathered}$$

Hence, the principal argument of $\mathit{z}\mathbf{=}\mathbf{3}\mathit{i}$ is $\mathit{\theta }\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}$.