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
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
$\displaystyle z=e^{i\theta} , \theta \in \mathbb{R}$ and $\displaystyle \theta$ is the angle measured from the positive x axis
For the numerator you have
$\displaystyle -4\sqrt{2} + 4\sqrt{2}i=4\sqrt{2}(-1+i)$
So lets just focus on the $\displaystyle -1+i$
$\displaystyle a+bi=e^{i\theta}=\sqrt{a^2+b^2}e^{i\tan^{-1}\left( \frac{b}{a}\right)}$
Also in the point is in the 2nd or 3rd quadrants you must at $\displaystyle \pi$ to the angle.
$\displaystyle -1+i=\sqrt{1^2+1^2}e^{i(\tan(\frac{-1}{1})+\pi)}=\sqrt{2}e^{i\frac{3\pi}{4}}$
Now if we recombine this with what we factored out it give
$\displaystyle 4\sqrt{2}\left(\sqrt{2}e^{i\frac{3\pi}{4}}\right)= 8e^{i\frac{3\pi}{4}}$
Now do the same thing to the denominator and then use exponential rules
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.