Question

convert the following complex number into its polar representation 2+2i

Answer 1

\[a+bi=r\left(\cos(\theta)+i\sin(\theta)\right)\] what you need is \(r\) and \(\theta\)
\[r=\sqrt{a^2+b^2}\]

Answer 2

and in this case (although not always), \(\theta=\tan^{-1}\left(\frac{b}{a}\right)\)

Answer 3

2\[2\sqrt{2}\]? then what

Answer 4

find \(\theta\) via \(\theta=\tan^{-1}\left(\frac{b}{a}\right)\)

Answer 5

or you can just find \(\theta\) in this case by thinking what the angle has to be

Answer 6

i got .7853 for tan

Answer 7

i would go for \(\frac{\pi}{4}\)

Answer 8

okay then what after that

Answer 9

then you are done

Answer 10

thank you i se

Answer 11

see

Answer 12

don't forget this equality is between numbers, one number equals another number

Answer 13

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

Answer 14

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