Does a purely imaginary number have a corresponding "angle" in polar coordinate system? Let's say we have a pure imaginary number with no real part, $i$.
I know that complex numbers in the form $a+bi$ can be converted into the polar coordinate system using the following relations:


*

*$\theta = \arctan{Im/Re} $

*$r = \sqrt{a^2+b^2} $


However, for a purely imaginary $i$ number with no real part, relation $1$ gives:
$$\theta = \arctan{1/0} $$
which is division by zero?
 A: The rule $\theta = \arctan \frac ba$ where $z = a+ib$ is wrong in multiple ways.
As you noticed, it does not correctly handle the case where $a = 0$ and $b \neq 0.$
It also gives incorrect answers for all numbers with negative real parts!
For example, let $z = -2 - i2.$ Then the arc tangent formula says
$$\theta = \arctan \frac {-2}{-2} = \arctan 1 = \frac\pi4,$$
but the correct answer is $\theta = \frac54\pi.$
In general, use the arc tangent formula only as a helpful hint to compute answers that are not obvious, and apply the obvious "fix" to the formula in the case where the real part is negative.
Remember, given $z = a + ib,$ the goal is to find $r$ and $\theta$ such that
\begin{align}
a &= r \cos\theta,\\
b &= r \sin\theta.
\end{align}
If a method achieves that result, use it; otherwise use a different method.
A: The way you define the phasor angle is inverse tangent. Just drop it and find the angle with i.e. inverse sine.
A: HINT:
We have $$a+ib=r(\cos\theta+i\sin\theta)$$
If $\displaystyle a=0,  \cos\theta=0\implies\sin\theta=\pm1 $
If $\displaystyle \sin\theta=1\iff b=r>0\implies \theta=\frac\pi2$
What if $\displaystyle \sin\theta=-1?$
Reference : The definition of arctan(x,y)
