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Express $ z=-32 $ in exponential form.

My reasoning:

  1. $ z=-32 $ is the same as $ z=-32+0i $

  2. Exponential form should look like $ z=Re^{\theta i} $

  3. $ R =\sqrt{(-32)^2 + 0^2} = 32 $

  4. $ \theta = \tan^{-1}(\frac{0}{-32}) = 0 $

  5. So answer becomes: $ z = 32e^{0i} = 32 $

But, obviously, $ -32 \ne 32 $

What am I missing?

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5 Answers

up vote 1 down vote accepted

Since $z=re^{i \theta}$, with $r\geq 0$, you must have $|z| = r$. So $r=32$. Then you need to solve $-32 = 32 e^{i \theta} = 32 (\cos\theta + i \sin \theta)$. Since the imaginary part is zero, you must have $\theta = n \pi$ for some integer $n$. Choosing $n$ to be odd gives all solutions, since $\cos n \pi = -1$ iff $n$ is odd.

Consequently $-32 = 32 e^{i n \pi}$, where $n$ is odd.

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$\tan^{-1}(0)$ could be both $0$ and $\pi$, and in your case, the second one fits.

$-32 = 32 e^{i\pi}$.

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The problem is with the inverse tangent that you're using to find the value of the argument of $-32$. Remember that the inverse tangent will give you values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ so it is not giving you the correct argument.

In this case since $-32$ is in the negative real axis, its argument should be $\pi$ instead. Which will give you the "exponential form"

$$ -32 = 32e^{\pi i} $$

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As $R\cos\theta=-32<0=>\cos\theta<0$ and $\sin\theta=0$ so, $\theta$ lie in the 3rd quadrant $\pi ≤ \theta ≤ 3\frac{\pi}{2}$,

and $tan^{-1}(0)=n\pi$ where n is any integer.

So here, $\theta=(2m+1)\pi$ where m is any integer.

So, the general value of $-32=32e^{i(2m+1)\pi}$, the principal value being $32e^{i\pi}$

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$ -32 = -32(1 + 0 i) = -32 e^{i2n\pi}$

Or, $32(-1 + 0 i) = 32( \cos \pi + i \sin \pi) = 32 e^{i \pi}$

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