SOLUTION: Express 2 - 2i in polar form using principal argument

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Question 1138879: Express 2 - 2i in polar form using principal argument
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Express 2+-+2i in polar form using principal argument
z=2+-+2i=>z=x+-+yi
The modulus (or length) of the complex number is simply given by using Pythagoras' theorem to find the hypotenuse, ie
z=r%2Acos%28theta%29%2Br%2Ai%2Asin%28theta%29
z=r%28cos%28theta%29%2Bi%2Asin%28theta%29%29
z=2+-+2i
r%2Acos%28theta%29=2
r%2Asin%28theta%29=-2
=>tan%28theta%29=-2%2F2=-1
will be easier to find acute angle in quadrant I whose tan is 1

alpha=tan%5E%28-1%29%281%29=pi%2F4 or alpha=45°
now find r
r=sqrt%282%5E2%2B%28-2%29%5E2%29
r=sqrt%288%29
r=sqrt%284%2A2%29
r=2sqrt%282%29
=> r is in quadrant IV
=> theta=360-alpha=360-45=315° or
theta=2pi-alpha=2pi-pi%2F4=7pi%2F4=> we can also choose to use theta=-alpha which is also in quadrant IV
so, theta=45° or theta=-pi%2F4
then we have
z=2sqrt%282%29%28cos%28-pi%2F4%29%2Bi%2Asin%28-pi%2F4%29%29......since cos%28-pi%2F4%29=cos%28pi%2F4%29 and sin%28-pi%2F4%29=+-sin%28pi%2F4%29, we have
z=2sqrt%282%29%28cos%28pi%2F4%29-i%2Asin%28pi%2F4%29%29
or usingtheta=45° , we have z=2sqrt%282%29%28cos%2845%29-i%2Asin%2845%29%29
also, using theta=7pi%2F4 or theta=315° :
z=2sqrt%282%29%28cos%287pi%2F4%29%2Bi%2Asin%287pi%2F4%29%29
or z=2sqrt%282%29%28cos%28315%29%2Bi%2Asin%28315%29%29