Write the complex number `z=-1-i` in polar form.
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.
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` .
`-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)`
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.
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
where a is the real part and b is the imaginary part of the complex number. So:
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:
An alternative is to write in Euler notation:
`z=|z|e^(i theta)=re^(i theta)`
So here we have:
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`