Question

express z=(sqrt3-i)^7/(sqrt2+isqrt2)^6 in trig form z=p(cos theta + i sin theta)

Answer 1

I not sure where to start should I convert the sqrt3-i into polar form?

Answer 2

Convert the numerator and denominator into polar separately.

Answer 3

\[\Large\rm \sqrt3-1i\]\[\Large\rm r=\sqrt{(\sqrt3)^2+(1)^2}=\sqrt{10}\]\[\Large\rm \tan \theta=\frac{-1}{\sqrt3}\implies\quad \theta=\frac{5\pi}{6}\]So in the numerator we have:\[\Large\rm \left(\sqrt{10}\left(\cos\frac{5\pi}{6}+i \sin\frac{5\pi}{6}\right)\right)^7\]From there we would apply De' Moivre's Theorem to deal with the exponent.
I skipped over a few steps in there, any confusion on finding the radial distance or the angle?

Answer 4

Woops, that r should be... 3+1, sqrt(4)=2.

Answer 5

that's what i got for the numerator =2, so , the denominator look like it's already in polar form

Answer 6

It does? 0_o hmm

Answer 7

So lemme wrap up the numerator a sec:\[\Large\rm =\left(2\left[\cos\frac{5\pi}{6}+i \sin\frac{5\pi}{6}\right]\right)^7\]
\[\Large\rm =2^7\left[\cos\frac{35\pi}{6}+i \sin\frac{35\pi}{6}\right]\]So this is our numerator in polar form ^
We would like to do similar to the denominator

Answer 8

Can you apply those same steps to the denominator?
We need a radius, and an angle.

Answer 9

..?
Confused?
What's going on in the brain there? :o

Answer 10

ohh no sorry computer was freezing lol , so denominator i got 2^6(cos 45deg(pi/4)+i sin 45deg)^6

Answer 11

then what would be the next step once both are in polar form

Answer 12

Good good good.
So apply De' Moivre's Theorem gets us to this part of the problem:\[\Large\rm =\frac{2^7\left(\cos\frac{35\pi}{6}+i \sin\frac{35\pi}{6}\right)}{2^6\left(\cos\frac{6\pi}{4}+i \sin\frac{6\pi}{4}\right)}\]

Answer 13

I can't remember what this rule is called.. but you're allowed to this:\[\Large\rm \frac{r_1(\cos a+i \sin a)}{r_2(\cos b+i \sin b)}=\frac{r_1}{r_2}\left[\cos(a-b)+i \sin(a-b)\right]\]We subtract the angles, and divide the radial lengths.

Answer 14

It follows from the fact that these polar expressions can be written as exponentials. And the exponential law that we're familiar with (dividing exponentials), carries over to complex numbers.

Answer 15

soo this formula can be used, i thought the r1/r2 formula was for two completly separate equations

Answer 16

If we converted to exponentials, it might make more sense why this rule can be applied:\[\Large\rm \frac{2^7\left(\cos\frac{35\pi}{6}+i \sin\frac{35\pi}{6}\right)}{2^6\left(\cos\frac{6\pi}{4}+i \sin\frac{6\pi}{4}\right)}=\frac{2^7e^{(35\pi /6)i}}{2^6e^{(6\pi/4)i}}\]And of course, our rules of exponents would tell us that this is:\[\Large\rm \frac{2^7}{2^6}e^{\left(\frac{35\pi}{6}-\frac{6\pi}{4}\right)i}\]

Answer 17

ohh ok makes sense when you use e

Answer 18

Shouldn't be too bad from there, right? :)
The 2's divide nicely.
Just need to find a common denominator so you can subtract.

Answer 19

yes thanks a lot

Answer 20

np