Question 1138879
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*cos(theta)+r*i*sin(theta)}}}
{{{z=r(cos(theta)+i*sin(theta))}}}

{{{z=2 - 2i}}}
{{{r*cos(theta)=2}}}
{{{r*sin(theta)=-2}}}
=>{{{tan(theta)=-2/2=-1}}}

will be easier to find acute angle in quadrant I whose tan is {{{1}}}


{{{alpha=tan^(-1)(1)=pi/4}}} or {{{alpha=45}}}°

now find {{{r}}}
{{{r=sqrt(2^2+(-2)^2)}}}
{{{r=sqrt(8)}}}
{{{r=sqrt(4*2)}}}
{{{r=2sqrt(2)}}}

=> {{{r}}} is in quadrant IV 
=> {{{theta=360-alpha=360-45=315}}}° or 
{{{theta=2pi-alpha=2pi-pi/4=7pi/4}}}=> we can also choose to use {{{theta=-alpha}}} which is also in quadrant IV

so, {{{theta=45}}}° or  {{{theta=-pi/4}}}

then we have

{{{z=2sqrt(2)(cos(-pi/4)+i*sin(-pi/4))}}}......since {{{cos(-pi/4)=cos(pi/4)}}} and {{{sin(-pi/4)= -sin(pi/4)}}}, we have

{{{z=2sqrt(2)(cos(pi/4)-i*sin(pi/4))}}} 
or using{{{theta=45}}}° , we have  {{{z=2sqrt(2)(cos(45)-i*sin(45))}}}

also, using {{{theta=7pi/4}}} or {{{theta=315}}}° :

{{{z=2sqrt(2)(cos(7pi/4)+i*sin(7pi/4))}}} 
or {{{z=2sqrt(2)(cos(315)+i*sin(315))}}}