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

1 Answer
Douglas K.
Apr 9, 2017

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

Explanation:

Given: #2+3i#

#r = sqrt(2^2+3^2)#

#r = sqrt(4+9)#

#r = sqrt(13)#

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

#theta = tan^-1(b/a)+0#

#theta = tan^-1(3/2)#

#theta ~~ 56.3^@ or 0.983" radians"#

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

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