# 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:

1. $\theta = \arctan{Im/Re}$
2. $r = \sqrt{a^2+b^2}$

However, for a purely imaginary $i$ number with no real part, relation $1$ gives:

$$\theta = \arctan{1/0}$$

which is division by zero?

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## 1 Answer

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)

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I suppose this is somewhat related to the limit of $arctan$ as it approaches +- $\infty$? – Bob Apr 8 '14 at 4:44