Question

y= -2+2i and z=1+i(squareroot)3
find yz in polar form

Answer 1

you can put them in polar form first, then multiply

Answer 2

do you know how to put them in polar form?

Answer 3

not really

Answer 4

you need two numbers, \(r=\sqrt{a^2+b^2}\) and \(\theta\) where \(\tan(\theta)=\frac{b}{a}\)

Answer 5

for \(-2+2i\) you have \(a=-2,b=2\) and so \(r=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\)

Answer 6

then \(\tan(\theta)=\frac{2}{-2}=-1\) and so because you are in quadrant 2, \(\theta =\frac{3\pi}{4}\)

Answer 7

oops typo there
\[-2+2i=2\sqrt{2}\left(\cos(\frac{3\pi}{3})+i\sin(\frac{3\pi}{4})\right)\]

Answer 8

i was wondering where that came from

Answer 9

yeah it was a mistake

Answer 10

now repeat the process for \(1+\sqrt{3}i\)

Answer 11

in this case \(a=1,b=\sqrt3\) and so
\[r=\sqrt{1^2+\sqrt{3}^2}=\sqrt{4}=2\]

Answer 12

also \(\tan(\theta)=\sqrt{3}\) and since you are in quadrant 1 you have \(\theta=\frac{\pi}{3}\)

Answer 13

gives
\[1+\sqrt{3}i=2\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)\]

Answer 14

the last part is easiest
to multiply the two numbers, add the angles and multiply the modulus (absolute value, \(r\))

Answer 15

okay thank you so much

Answer 16

yw