How do you write the trigonometric form of #-2-2i#?

1 Answer
Nov 7, 2016

The trigonometric form is #z=2sqrt2(cos((5pi)/4)+isin((5pi)/4))#

Explanation:

Let #z=-2-2i#
The modulus of z is # |z|=sqrt(2^2+2^2)=2sqrt2 #
Then we rewrite z as #z=2sqrt2(-2/(2sqrt2)-2/(2sqrt2)i)#
simplify #z=2sqrt2(-sqrt2/2-sqrt2/2i)#
Comparing this to #z=r(costheta+isintheta)# which is the trgonometric form.
So #costheta=-sqrt2/2# and #sintheta=-sqrt2/2#
#:. theta# is in the 3rd quadrant
#theta=(5pi)/4#
So the trigonometric form is #z=2sqrt2(cos((5pi)/4)+isin((5pi)/4))

and the exponential form is #z=2sqrt2e^((5ipi)/4)#