How do you write the complex number in trigonometric form 3+sqrt3i3+√3i?

1 Answer
Narad T.
Oct 17, 2016

=2sqrt3(cos(pi/6)+sin(pi/6))=2√3(cos(π6)+sin(π6))

Explanation:

let z=3+isqrt3z=3+i√3
We have to calculate the modulus ∣z∣=sqrt(3^2+(sqrt3)^2)=sqrt(9+3)=sqrt12=2sqrt3∣∣∣z∣=√32+(√3)2=√9+3=√12=2√3
Now we have to divide z by ∣z∣
z/(∣z∣)z∣z∣=3/(2sqrt3)+isqrt3/(2sqrt3)=sqrt3/2+i/232√3+i√32√3=√32+i2
So z=2sqrt3(sqrt3/2+i/2)z=2√3(√32+i2)
Now you compare this to
z=r(costheta+isintheta )z=r(cosθ+isinθ)
You can see that
costheta=sqrt3/2cosθ=√32
and sintheta=1/2sinθ=12
As the signs are positive, the results is in the first quadrant
theta=30º orpi/6
So the anwer is z=2sqrt3(cos(pi/6)+sin(pi/6))

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