Write the complex number z=-1-i in polar form.

We are asked to write the complex number z=-1-i in polar form.
The polar form of a complex number is 
z=r(cos theta + i sin theta)
where r is the distance from the origin (the modulus or absolute value of the complex number) and theta is the angle from the positive x-axis. (Note that the angle is not unique -- you can always select an angle from 0 to 2pi or 0 to 360 degrees.)
To compute r we use 
r=sqrt(a^2+b^2)
where a is the real part and b is the imaginary part of the complex number. So:
r=sqrt((-1)^2+(-1)^2)=sqrt(2)
We can find theta by
tan theta = b/a
In this case we can solve the angle by inspection as 225 degrees or 5pi/4.
The polar form is:
z=sqrt(2)(cos((5pi)/4)+isin((5pi)/4))
An alternative is to write in Euler notation:
z=|z|e^(i theta)=re^(i theta)
So here we have:
-1-i=sqrt(2)e^((i(5pi)/4))
Note again that the angle is not unique; we can add/subtract any multiples of 2pi and still have the same point. A general solution is:
-1-i=sqrt(2)(cos((5pi)/4 + 2pi n)+isin((5pi)/4+ 2pi n))=sqrt(2)e^(i((5pi)/4+2pi n)); n in ZZ
 
http://mathworld.wolfram.com/ComplexNumber.html

The complex number z has the form z=x+iy . In the complex plane where x is the real axis and y is the imaginary axis we have x=-1 and y=-1 .
Find the hypotenuse.
r=sqrt((-1)^2+(-1)^2)=sqrt(2)
There are infinite values for theta that can make this triangle in the third quadrant.
theta=(5pi)/4+2pi n, n=0,+-1,+-2...
Now use the polar relations for x and y .
x+iy=[rcos(theta)]+i[r sin(theta)]
-1-i=sqrt(2)cos((5pi)/4+2pi n)+i sqrt(2) sin((5pi)/4+2pi n)
-1-i=sqrt(2)[cos((5pi)/4+2pi n)+i sin((5pi)/4+2pi n)]
Use Euler's relation
rcos(theta)]+i r sin(theta)=r*e^(i*theta)
-1-i=sqrt(2)*e^((5pi)/4+2pi n)
Therefore, z is a point on the circle sqrt(2)e^(i*theta) where theta has the angle (5pi)/4 which also works for every revolution of 2pi around the circle.
http://mathworld.wolfram.com/EulerFormula.html