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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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The way you define the phasor angle is inverse tangent. Just drop it and find the angle with i.e. inverse sine.

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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.

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

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