Can anyone help me with this problem?

Use polar form find the quotient express the result in rectangular form

(-4\sqrt{2} + 4\sqrt{2i}) divided by (6+6i)

So far I got

-.94-.94i+.94i+.94i^2 But am I doing this right(Wink)

Printable View

- Apr 28th 2011, 03:56 PMhomeylova223Quotient of complex numbers in polar form?
Can anyone help me with this problem?

Use polar form find the quotient express the result in rectangular form

(-4\sqrt{2} + 4\sqrt{2i}) divided by (6+6i)

So far I got

-.94-.94i+.94i+.94i^2 But am I doing this right(Wink) - Apr 28th 2011, 04:08 PMpickslides
mulitply through by (6-6i)/(6-6i)

- Apr 28th 2011, 04:12 PMTheEmptySet
First you have not written the numbers in polar form and I and going to assume that the i in the last term of the numerator is NOT under the radical.

A complex number is in polar form if

and is the angle measured from the positive x axis

For the numerator you have

So lets just focus on the

Also in the point is in the 2nd or 3rd quadrants you must at to the angle.

Now if we recombine this with what we factored out it give

Now do the same thing to the denominator and then use exponential rules:) - Apr 28th 2011, 04:25 PMhomeylova223
I might sound stupid but I actually wrote the problem from my textbook

(-4sqrt{2} + 4sqrt{2}i) divided by (6+6i)

I am sorry for writing it wrong. - Apr 28th 2011, 04:28 PMTheEmptySet
- Apr 28th 2011, 04:34 PMpickslides
I think I misread your post, in my initial post the idea was to make the quoitent into the form a+bi, then you can change this into polar form.

TheEmptySet's solution answers the question directly.

Both should arrive at the same answer, you should do both to prove this, it will increase your understanding of complex numbers. - Apr 28th 2011, 04:48 PMPlato
- Apr 28th 2011, 04:57 PMtopsquark
- Apr 28th 2011, 05:54 PMPlato
- Apr 29th 2011, 02:42 PMhomeylova223
- Apr 29th 2011, 03:06 PMPlato
- Apr 29th 2011, 04:16 PMhomeylova223
Actually I think Plato and I have a similar answer as mine can be reduced to 2 square root 2/3 i +0