Question 1150572
<pre>The polar form of A+Bi is 

{{{r(cos(theta)^""+i*sin(theta)^"")}}}

The complex number A+Bi is a vector (arrow) drawn from the point (0,0) to
the point (A,B).

So
The complex number -1-1i is a vector (arrow) drawn from the point (0,0) to
the point (-1,-1), which is this:

{{{drawing(400,400,-1.9,1.9,-1.9,1.9,

locate(-1.52,-1,"(-1,-1)"), locate(.025,.16,"(0,0)"),
grid(1), line(0,0,-1,-1),line(-.92,-.8,-1,-1),line(-.8,-.9,-1,-1) )}}}

We draw the right triangle which has one leg on the x-axis and the other leg
(in green) perpendicular to the x-axis and ending at the point (-1,-1):

{{{drawing(400,400,-1.9,1.9,-1.9,1.9,
green(line(0,0,-1,0),line(-1,0,-1,-1)),
locate(-1.52,-1,"(-1,-1)"), locate(.025,.16,"(0,0)"),
grid(1), line(0,0,-1,-1),line(-.92,-.8,-1,-1),line(-.8,-.9,-1,-1) )}}}

Then we calculate r, which is the length of the arrow, by the Pythagorean
theorem:

{{{r=sqrt(x^2+y^2)=sqrt((-1)^2+(-1)^2)=sqrt(1^""+1^"")=sqrt(2)}}}

Then we find θ measured counter-clockwise and swung around from the right
side of the x-axis, indicated by the red arc:

{{{drawing(400,400,-1.9,1.9,-1.9,1.9,
green(line(0,0,-1,0),line(-1,0,-1,-1)),

red(arc(0,0,1,-1,0,225)),
locate(-1.52,-1,"(-1,-1)"), locate(.025,.16,"(0,0)"),
grid(1), line(0,0,-1,-1),line(-.92,-.8,-1,-1),line(-.8,-.9,-1,-1) )}}}

Use the unit circle to find θ since it is a special angle.

So {{{theta=225^o=(5pi)/4}}}

The polar form for -1-i is 

{{{r(cos(theta)+i*sin(theta^""))}}}

{{{sqrt(2)(cos(225^o)^""+i*sin(225^o)^"")}}}

if your teacher wants the angle in degrees, or

{{{sqrt(2)(cos((5pi)/4)^""+i*sin((5pi)/4)^"")}}}

if your teacher wants the angle in radians.

Edwin</pre>