Question 152493
How would I write the complex # y= -2+2i in trig form?
<pre><font size = 4><b>
To draw the complex number {{{x+yi}}}, plot the point
(x,y) and connect it to the origin ({{{0}}},{{{0}}}).

So to draw the complex number {{{-2+2i}}}, we have {{{x=-2}}}
and {{{y=2}}}

Now we plot the point ({{{-2}}},{{{2}}}) and connect it to
the origin ({{{0}}},{{{0}}}).

{{{drawing(400,400,-3,3,-3,3,graph(400,400,-3,3,-3,3),line(-2,2,0,0),
locate(-2.2,2.2,"(-2,2)"))}}}

now we draw a perpendicular down to the x-axis:

{{{drawing(400,400,-3,3,-3,3,graph(400,400,-3,3,-3,3),line(-2,2,0,0),
line(-2,2,-2,0),locate(-2.2,2.2,"(-2,2)") ) }}}

We label {{{x=-2}}} and {{{y=2}}}

{{{drawing(400,400,-3,3,-3,3,graph(400,400,-3,3,-3,3),line(-2,2,0,0),
line(-2,2,-2,0), locate(-2.5,1,"y=2"), locate(-1.3,.3,"x=-2"),locate(-2.2,2.2,"(-2,2)") )}}}

Now we calculate r by the Pythagorean equation:

{{{r=sqrt(x^2+y^2)}}}

{{{r=sqrt((-2)^2+(2)^2)}}}

{{{r=sqrt(4+4)}}}

{{{r=sqrt(8)}}}

{{{r=sqrt(4*2)}}}

{{{r=sqrt(4)sqrt(2)}}}

{{{r=2sqrt(2)}}}

So we label {{{r}}} as {{{r=2sqrt(2)}}}

{{{drawing(400,400,-3,3,-3,3,graph(400,400,-3,3,-3,3),line(-2,2,0,0),
line(-2,2,-2,0), locate(-2.5,1,"y=2"), locate(-1.3,.3,"x=-2"),
locate(-.8,1.5,sqrt(2)), locate(-1.2,1.46,"r=2"),locate(-2.2,2.2,"(-2,2)")
 )}}}

Next we indicate the angle @ by a curved line:

{{{drawing(400,400,-3,3,-3,3,graph(400,400,-3,3,-3,3,sqrt(.36-x^2)*sqrt(x+.424261)/sqrt(x+.424261)),line(-2,2,0,0),
line(-2,2,-2,0), locate(-2.5,1,"y=2"), locate(-1.3,.3,"x=-2"),
locate(-.8,1.5,sqrt(2)), locate(-1.2,1.46,"r=2"),locate(.1,.8,"@"),locate(-2.2,2.2,"(-2,2)"))}}}

Now we determine the angle @ by any trig function

sin(@) = {{{y/r}}}, cos(@) = {{{x/r}}} or tan(@) = {{{y/x}}}

If we use the last one we have

tan(@) = {{{y/x = 2/(-2) = -1}}}

and since we know that 45° has tangent 1, we know
the reference angle is 45°, and since @ is in the
second quadrant, the actual angle is 180°-45°or 135°.

{{{drawing(400,400,-3,3,-3,3,graph(400,400,-3,3,-3,3,sqrt(.36-x^2)*sqrt(x+.424261)/sqrt(x+.424261)),line(-2,2,0,0),
line(-2,2,-2,0), locate(-2.5,1,"y=2"), locate(-1.3,.3,"x=-2"),
locate(-.8,1.5,sqrt(2)), locate(-1.2,1.46,"r=2"),locate(.1,.8,"@=135°"),locate(-2.2,2.2,"(-2,2)"))}}}

The trig form is 

r[cos(@) + i·sin(@)]

so we substitute and get:

{{{2sqrt(2)}}}[cos(135°) + i·sin(135°)]

and sometimes this is abbreviated as

{{{2sqrt(2)}}}cis(135°)

Edwin</pre>