Quotient of complex numbers in polar form?

**homeylova223** · April 28th 2011, 02:56 PM

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

**pickslides** · April 28th 2011, 03:08 PM

mulitply through by (6-6i)/(6-6i)

**TheEmptySet** · April 28th 2011, 03:12 PM

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

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

**homeylova223** · April 28th 2011, 03:25 PM

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.

**TheEmptySet** · April 28th 2011, 03:28 PM

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

**pickslides** · April 28th 2011, 03:34 PM

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.

**Plato** · April 28th 2011, 03:48 PM

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.

**topsquark** · April 28th 2011, 03:57 PM

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

**Plato** · April 28th 2011, 04:54 PM

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 $\overline{w}=r\exp(-\mathif{i}\theta)$ .

**homeylova223** · April 29th 2011, 01:42 PM

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?

**Plato** · April 29th 2011, 02:06 PM

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)

$\frac{{ - 4\sqrt 2 + 4\sqrt 2 i}}{{6 + 6i}} = \frac{{4\sqrt 2 e^{\frac{{3\pi i}}{4}} }}{{6e^{\frac{{\pi i}}{4}} }} = \frac{{2\sqrt 2 }}{3}e^{\frac{{\pi i}}{2}} =0+ \frac{{2\sqrt 2 }}{3}i$

**homeylova223** · April 29th 2011, 03:16 PM

Actually I think Plato and I have a similar answer as mine can be reduced to 2 square root 2/3 i +0

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