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Find $zw, \frac{z}{w},\frac{1}{z}$ for $ z=2\sqrt{3}-2i, w=-1+i$

I went wrong somewhere, this is what I have so far (this is in polar):

$z=4\left(\cos\left(\frac{11\pi}{6}\right)+\sin\left(\frac{11\pi}{6}\right)\right) $

$w=\sqrt2\left(\cos\left(\frac{7\pi}{4}\right)+\sin\left(\frac{7\pi}{4}\right)\right) $

Now my setup should be:

$zw=4\sqrt2\left(\cos\left(\frac{11\pi}{6}+\frac{7\pi}{4}\right)+\sin\left(\frac{11\pi}{6}+\frac{7\pi}{4}\right)\right) $

The common denominator is $12$ so

$zw=4\sqrt2\left(\cos\left(\frac{22\pi}{12}+\frac{21\pi}{12}\right)+\sin\left(\frac{22\pi}{12}+\frac{21\pi}{12}\right)\right) $

which then should equal out to

$zw=4\sqrt2\left(\cos\left(\frac{43\pi}{12}\right)+\sin\left(\frac{43\pi}{12}\right)\right) $

The answer in the book says:

$zw=4\sqrt2\left(\cos\left(\frac{7\pi}{12}\right)+\sin\left(\frac{7\pi}{12}\right)\right) $

Where did I go wrong?

I haven't even tried the other problems yet.

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  • $\begingroup$ What do you mean by find zw, z/w, and 1/z? you want the result of the form a+ib? $\endgroup$ – Jika Jun 23 '14 at 20:35
  • $\begingroup$ @jika zw is a form, z1z2. complex numbers of polar form. $\endgroup$ – Joshhw Jun 23 '14 at 20:37
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First, notice that the argument of $w$ is $\frac{3 \pi}{4}$, not $\frac{7\pi}{4}$. And you forgot to put the "$i$"'s together with the sines. Just a little distraction. The other argument, and the absolute values are ok. You had setup everything else correctly. Using the right value above, we get: $$\begin{align}zw &=4\sqrt2\left(\cos\left(\frac{11\pi}{6}+\frac{3\pi}{4}\right)+i\sin\left(\frac{11\pi}{6}+\frac{3\pi}{4}\right)\right) \\ &= 4\sqrt2\left(\cos\left(\frac{22\pi + 9\pi}{12}\right)+i\sin\left(\frac{22\pi + 9\pi}{12}\right)\right) \\ &= 4\sqrt2\left(\cos\left(\frac{31\pi}{12}\right)+i\sin\left(\frac{31\pi}{12}\right)\right)\end{align}$$ Normally, we would stop here, but, we can always reduce the argument so it is between $0$ and $2\pi$, and use that $\sin$ and $\cos$ have both period $2\pi$. Notice that: $$\frac{31\pi}{12} = \frac{7\pi}{12} + \frac{24\pi}{12} = \frac{7\pi}{12} + 2\pi$$ This way, we obtain: $$zw = 4\sqrt2\left(\cos\left(\frac{7\pi}{12}\right)+i\sin\left(\frac{7\pi}{12}\right)\right)$$ as desired. Ok?

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  • $\begingroup$ how is it $\frac{3\pi}{4}$? is it because the circle goes counterclockwise and I should've found the that angle before the one I used? $\endgroup$ – Joshhw Jun 23 '14 at 22:21
  • $\begingroup$ Plot the point in the complex plane to see it better. Yes, we count it counterclockwise. The point in the line $y = -x$. $\endgroup$ – Ivo Terek Jun 23 '14 at 22:23
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You have the angle wrong for $w$. It should be $3\pi/4$, right?

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