Expressing complex numbers in polar form.

**rorosingsong** · March 25th 2011, 01:03 AM

Hi, I'm not too sure with this one, can I get some help?

If Z= 3 - 3i;what are Z, 1/Z and $Z^{4}$ in polar form?

I'm not sure if my answers are correct...

Thanks, any help would be greatly appreciated!

Cheers.

**Prove It** · March 25th 2011, 01:04 AM

Well, what are your answers?

**rorosingsong** · March 25th 2011, 03:13 AM

Mmmm... I got:

$z = 3\sqrt{2}\left ( cos(-\frac{\pi }{4}) + i sin(-\frac{\pi }{4})\right )$

$z^{4} = 324\left ( cos(\pi) + i sin(\pi)\right )$

and

$\frac{1}{z} = \frac{\sqrt{2}}{6} ( cos(\frac{\pi }{4}) + i sin(\frac{\pi }{4})\right )$

**mr fantastic** · March 25th 2011, 03:32 AM

Originally Posted by rorosingsong

Mmmm... I got:

$z = 3\sqrt{2}\left ( cos(-\frac{\pi }{4}) + i sin(-\frac{\pi }{4})\right )$ Mr F says: Correct.

$z^{4} = 324\left ( cos(\pi) + i sin(\pi)\right )$ Mr F says: Argument is wrong. How did you get it?

and

$\frac{1}{z} = \frac{\sqrt{2}}{6} ( cos(\frac{\pi }{4}) + i sin(\frac{\pi }{4})\right )$ Mr F says: Correct.

..

**rorosingsong** · March 25th 2011, 11:27 PM

Oh whoops... silly mistake, methinks.

$z^{4} = 324 (cos(-\pi )+ i sin(-\pi ))$

Is that right?

**Prove It** · March 25th 2011, 11:28 PM

Yes

**HallsofIvy** · March 26th 2011, 12:33 PM

Originally Posted by rorosingsong

Oh whoops... silly mistake, methinks.

$z^{4} = 324 (cos(-\pi )+ i sin(-\pi ))$

Is that right?

Of course, adding $2\pi$ to the argument does't change the value but gives $324(cos(\pi)+ i sin(\pi))$ , exactly what you had before.

**Prove It** · March 26th 2011, 06:39 PM

Originally Posted by HallsofIvy

Of course, adding $2\pi$ to the argument does't change the value but gives $324(cos(\pi)+ i sin(\pi))$ , exactly what you had before.

True, but it is not the correct application of DeMoivre's Theorem.

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