Questions on Algebra: Complex Numbers answered by real tutors!

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Question 152493: how would i write the complex # y= -2+2i in trig form?: how would i write the complex # y= -2+2i in trig form?
Answer by Edwin McCravy(2086) About Me  (Show Source):
You can put this solution on YOUR website!
How would I write the complex # y= -2+2i in trig form?

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