How do you write 2+3i2+3i in trigonometric form?

1 Answer
Apr 9, 2017

To convert, a+bitor(cos(theta)+isin(theta))a+bir(cos(θ)+isin(θ)), use:
r=sqrt(a^2+b^2)r=a2+b2
theta=tan^-1(b/a)+0,pi,pi,2piθ=tan1(ba)+0,π,π,2π
Depending on whether the signs of "a" and "b" indicate the, 1st, 2nd, 3rd or 4th quadrant.

Explanation:

Given: 2+3i2+3i

r = sqrt(2^2+3^2)r=22+32

r = sqrt(4+9)r=4+9

r = sqrt(13)r=13

The signs of "a" and "b" indicate the 1st quadrant:

theta = tan^-1(b/a)+0θ=tan1(ba)+0

theta = tan^-1(3/2)θ=tan1(32)

theta ~~ 56.3^@ or 0.983" radians"θ56.3or0.983 radians

2+3i to sqrt(13)(cos(56.3^@)+isin(56.3^@))2+3i13(cos(56.3)+isin(56.3))