How do I find the trigonometric form of the complex number -1-isqrt3−1−i√3?

1 Answer
Gió
Jan 18, 2015

This complex number can be represented graphically as:enter image source here
To represent it in trigonometric form I first need to calculate the modulus rr using Pitagora's theorem:
r=sqrt((-1)^2+(-sqrt(3))^2)=sqrt(1+3)=2r=√(−1)2+(−√3)2=√1+3=2
Then I need to find the angle thetaθ;
theta=pi+arctan(sqrt(3))=4/3piθ=π+arctan(√3)=43π
So:
z=-1-isqrt(3)=2[cos(4/3pi)+isin(4/3pi)]z=−1−i√3=2[cos(43π)+isin(43π)]

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