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.

| cite | improve this question | | | | |
  • $\begingroup$ 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

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?

| cite | improve this answer | | | | |
  • $\begingroup$ 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

You have the angle wrong for $w$. It should be $3\pi/4$, right?

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.