Quotient of complex numbers in polar form?

• Apr 28th 2011, 02:56 PM
homeylova223
Quotient 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, 03:08 PM
pickslides
mulitply through by (6-6i)/(6-6i)
• Apr 28th 2011, 03:12 PM
TheEmptySet
Quote:

Originally Posted by homeylova223
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)

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
$z=e^{i\theta} , \theta \in \mathbb{R}$ and $\theta$ is the angle measured from the positive x axis

For the numerator you have

$-4\sqrt{2} + 4\sqrt{2}i=4\sqrt{2}(-1+i)$

So lets just focus on the $-1+i$

$a+bi=e^{i\theta}=\sqrt{a^2+b^2}e^{i\tan^{-1}\left( \frac{b}{a}\right)}$

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

$-1+i=\sqrt{1^2+1^2}e^{i(\tan(\frac{-1}{1})+\pi)}=\sqrt{2}e^{i\frac{3\pi}{4}}$

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

$4\sqrt{2}\left(\sqrt{2}e^{i\frac{3\pi}{4}}\right)= 8e^{i\frac{3\pi}{4}}$

Now do the same thing to the denominator and then use exponential rules:)
• Apr 28th 2011, 03:25 PM
homeylova223
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, 03:28 PM
TheEmptySet
Quote:

Originally Posted by homeylova223
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.

It is fine don't worry about it. I figured as much my above post still holds try to write the denominator in polar form! :)

Good luck
• Apr 28th 2011, 03:34 PM
pickslides
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, 03:48 PM
Plato
Quote:

Originally Posted by homeylova223
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.

Do you know that for any two complex numbers this is true?
$\frac{z}{w}=\frac{z\overline{w}}{|w|^2}$

That simple fact makes these questions trivial.
• Apr 28th 2011, 03:57 PM
topsquark
Quote:

Originally Posted by Plato
Do you know that for any two complex numbers this is true?
$\frac{z}{w}=\frac{z\overline{w}}{|w|^2}$

That simple fact makes these questions trivial.

True, but the OP says to use polar form to find the answer. However your formula would provide an easy check for the answer.

-Dan
• Apr 28th 2011, 04:54 PM
Plato
Quote:

Originally Posted by topsquark
True, but the OP says to use polar form to find the answer. However your formula would provide an easy check for the answer.

But that is even easier.
If $w=r\exp(\mathif{i}\theta)$ then $\overline{w}=r\exp(-\mathif{i}\theta)$.
• Apr 29th 2011, 01:42 PM
homeylova223
Quote:

Originally Posted by pickslides
mulitply through by (6-6i)/(6-6i)

I did that and now I get 48 square root 2 i / 72

Is there any way I could get it to a+bi form now?
• Apr 29th 2011, 02:06 PM
Plato
Quote:

Originally Posted by homeylova223
Use polar form find the quotient express the result in rectangular form
(-4\sqrt{2} + 4\sqrt{2i}) divided by (6+6i)

$\frac{{ - 4\sqrt 2 + 4\sqrt 2 i}}{{6 + 6i}} = \frac{{4\sqrt 2 e^{\frac{{3\pi i}}{4}} }}{{6e^{\frac{{\pi i}}{4}} }} = \frac{{2\sqrt 2 }}{3}e^{\frac{{\pi i}}{2}} =0+ \frac{{2\sqrt 2 }}{3}i$
• Apr 29th 2011, 03:16 PM
homeylova223
Actually I think Plato and I have a similar answer as mine can be reduced to 2 square root 2/3 i +0