# Expressing complex numbers in polar form.

• Mar 25th 2011, 01:03 AM
rorosingsong
Expressing complex numbers in polar form.
Hi, I'm not too sure with this one, can I get some help?

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

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

Thanks, any help would be greatly appreciated!

Cheers.
• Mar 25th 2011, 01:04 AM
Prove It
• Mar 25th 2011, 03:13 AM
rorosingsong
Mmmm... I got:

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

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

and

$\displaystyle \frac{1}{z} = \frac{\sqrt{2}}{6} ( cos(\frac{\pi }{4}) + i sin(\frac{\pi }{4})\right )$
• Mar 25th 2011, 03:32 AM
mr fantastic
Quote:

Originally Posted by rorosingsong
Mmmm... I got:

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

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

and

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

..
• Mar 25th 2011, 11:27 PM
rorosingsong
Oh whoops... silly mistake, methinks.

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

Is that right?
• Mar 25th 2011, 11:28 PM
Prove It
Yes
• Mar 26th 2011, 12:33 PM
HallsofIvy
Quote:

Originally Posted by rorosingsong
Oh whoops... silly mistake, methinks.

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

Is that right?

Of course, adding $\displaystyle 2\pi$ to the argument does't change the value but gives $\displaystyle 324(cos(\pi)+ i sin(\pi))$, exactly what you had before.
• Mar 26th 2011, 06:39 PM
Prove It
Quote:

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

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