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

1 Answer
Narad T.
Nov 7, 2016

The trigonometric form is z=2sqrt2(cos((5pi)/4)+isin((5pi)/4))z=2√2(cos(5π4)+isin(5π4))

Explanation:

Let z=-2-2iz=−2−2i
The modulus of z is |z|=sqrt(2^2+2^2)=2sqrt2 |z|=√22+22=2√2
Then we rewrite z as z=2sqrt2(-2/(2sqrt2)-2/(2sqrt2)i)z=2√2(−22√2−22√2i)
simplify z=2sqrt2(-sqrt2/2-sqrt2/2i)z=2√2(−√22−√22i)
Comparing this to z=r(costheta+isintheta)z=r(cosθ+isinθ) which is the trgonometric form.
So costheta=-sqrt2/2cosθ=−√22 and sintheta=-sqrt2/2sinθ=−√22
:. 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)

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1873 views around the world
You can reuse this answer
Creative Commons License