How do you find the trigonometric form of the complex number 3i?

1 Answer
Dec 21, 2014

#z=3[cos(pi/2)+isin(pi/2)]=3isin(pi/2)#

When you have to convert a complex number, given in "rectangular form" ( #z=a+ib# ), to trigonometric form #z=r[cos(theta)+isin(theta)]# you need to evaluate:
1) the modulus #r# (using Pitagora's Theorem);
2) the argument #theta# (using trigonometry).

Graphically:
enter image source here

In your case you have: #z=0+3i=3i# so that:
1) #r=sqrt(3^2+0^2)=3#
2) #theta=arctan(3/0)=pi/2#

Graphically:
enter image source here