Question

Determine the polar form of the complex number 2-2i. Express the angle theta in degrees, where 0 <= theta <= 360, and round numerical entries in the answer to two decimal places.

Answer 1

you need two numbers to write \[2-2i\] as \[r\left(\cos(\theta)+i\sin(\theta)\right)\] namely \(r\) and \(\theta\)

Answer 2

\(r\) is the easiest, it is \[r=\sqrt{a^2+b^2}\]

Answer 3

\[r=\sqrt{(2)^2+(-2i)^2}\]correct?

Answer 4

no

Answer 5

in \(a+bi\) both \(a\) and \(b\) are real numbers

Answer 6

oh..so it's just \[\sqrt{2^2+(-2)^2}\]?

Answer 7

so it is just \[\sqrt{2^2+2^2}\]
no \(i\)'s involved
it is the distance the number is from \((0,0)\) in the complex plane
yeah what you wrote

Answer 8

okay..

Answer 9

so sqrt8?

Answer 10

as for \(\theta\) that is pretty easy in this case if you think about where the number is

Answer 11

yeah \(\sqrt8\) or maybe \(2\sqrt2\)

Answer 12

It will be in decimals so 2.83 XD

Answer 13

oh no

Answer 14

leave it as \[2\sqrt2\]

Answer 15

Okay...

Answer 16

can you tell what \(\theta\) is from the picture?

Answer 17

Not really....perhaps 315 degrees? >.<

Answer 18

yup

Answer 19

both sides are the same, so it is like that 45 - 45 - 90 right triangle, so you got it, \(315^\circ\) if you are working in degrees

Answer 20

awesome

Answer 21

you could also say \(-45^\circ\)

Answer 22

But, the problem has a state it in it's positive form XD

Answer 23

us*

Answer 24

make sure this is clear, it is an equality between numbers \[2-2i=2\sqrt2\left(\cos(315)+i\sin(315)\right)\]

Answer 25

the number on the left is equal to the number on the right
if you evaluate the trig functions and multiply both by \(2\sqrt2\) you will get \(2-2i\) for sure

Answer 26

oh