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I'm trying to change the complex number, $-3i$ to polar representation.

What I did:

$a=0$
$b=-3$

$r=\sqrt{a^2+b^2} = \sqrt{0+3^2} = 3$

$\theta = \frac{b}{a} = \frac{-3}{0}$

But after that I'm stuck, since $\frac{-3}{0}$ is undefined. Where did I go wrong?

Yiyuan Lee
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Tyler
  • 21

1 Answers1

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$\tan(\theta) = \frac{b}{a}$ is the correct form to determine $\theta$. Though, for this problem, there is no need to use this formula to represent $-3i$ in polar value. Keep in mind that $\theta$ is between $-\pi$ and $\pi$. That is the principal argument of the complex number, denoted by $\mathrm{Arg}$. Since the imaginary part of the given complex number is nonzero and real part is zero, we conclude that $\mathrm{Arg}(-3i) = -\frac{\pi}{2}$. Thus, we have

$$-3i = 3\left(\cos\left(-\dfrac{\pi}{2} \right) + i\sin\left(-\dfrac{\pi}{2} \right)\right) = 3e^{-\frac{\pi}{2}i}$$

Another way to write this in polar representation is to look at $-i$. Then,

$$-i = \cos\left(-\dfrac{\pi}{2} \right) + i\sin\left(-\dfrac{\pi}{2} \right) = e^{-\frac{\pi}{2}i}$$

Multiply that by $3$ to get the answer. You can determine $\theta$ and $r$, using the equations you indicated, but for this type of problem you don't need to. If you observe the real and imaginary parts of the complex number carefully, then you should quickly identify $r$ and $\theta$ without efforts.

NasuSama
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  • Would this be raised to the positive π/2? Since b was negative? – Tyler Mar 09 '14 at 02:18
  • No since I determine the principal argument of the complex number. As mentioned in my answer, $-\pi < \mathrm{Arg} \leq \pi$. Also, since radius is nonnegative, we need $\theta = -\frac{\pi}{2}$, so by Euler's identity we have negative radians as the exponent of $e$. – NasuSama Mar 09 '14 at 02:20
  • Makes sense. I'm guessing my book must have made an error. Their answer was: 3cis(3π/2) – Tyler Mar 09 '14 at 02:24