Question

Ok I have worked this out but apparently I am using the incorrect angle. Find the product of complex numbers. leave in polar form. z1=2+2i z2=sqrt3 - i. I get 2(cos5pi/6+isin5pi/6)2(cos5pi/4+isin5pi/4). steps i took were 1)r=sqrt x^2+y^2=2sqrt2 2) tan^-1(theta)=pi/4 quad 1=45degrees. pi/4+pi=5pi/4 3) plug into get 2sqrt2(cos5pi/4 +isin5pi/4) 4) z2=rsintheta/rcostheta etc...

Answer 1

wow that is like impossible to read

Answer 2

\[2+2i=2\sqrt2(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))\] right?

Answer 3

new here. how can i make it better?

Answer 4

it's ok, is what i wrote correct?

Answer 5

maybe a mistake in the second one
\[\sqrt3-i\]

Answer 6

yeah i know i am just trying to suss out what he/she wrote after the first line

Answer 7

so z1 is 2+2i
z2 is sqrt3 -i

Answer 8

polar form of \[\sqrt3-i\] is the question
in this case \(r=2\) right

Answer 9

and you have choices for \(\theta\) as always

Answer 10

i would pick \(\theta=-\frac{\pi}{6}\)

Answer 11

yes it needs to be answered in polar form after taking the product of. so yes the choices in angles I think may be my big problem.

Answer 12

what did you pick?

Answer 13

should be either
\[-\frac{\pi}{6}\] or
\[\frac{11\pi}{6}\]

Answer 14

5pi/4..soo i am all messed up. suppose we can take it from the top? oh you are doing \[\sqrt{3}-i\], yes i chose -pi/6 then converted to get 5pi/6

Answer 15

you cannot convert like that

Answer 16

you can go around the unit circle, but not across it

Answer 17

\[2\sqrt2(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))\times 2(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6}))\] multiply the absolute values, add the angles

Answer 18

ok then i get 5pi/12 but the answer sheet says 4sqrt2(cos pi/12+isin pi/12)

Answer 19

What is a good system for choosing an angle?