How do you convert #9i # to polar form?

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Answer 1 · 2018-04-11T09:15:45

#color(blue)(9[cos(pi/2)+isin(pi/2)]#

For complex numbers given in the form:

#a+bi#

The polar form will be:

#z=r[cos(theta)+isin(theta)]#

Where:

#r=sqrt(a^2+b^2)#

#theta=arctan(b/a)#

#0+9i#

#r=sqrt((0)^2+(9)^2)=9#

#theta=arctan(9/0)#

This is undefined, which tells us that #theta# is either #pi/2# or #(5pi)/2#.

#9i# is positive, so this must be in the I and II quadrants. #theta# is therefore #pi/2#

Plugging these values into:

#z=r[cos(theta)+isin(theta)]#

#z=color(blue)(9[cos(pi/2)+isin(pi/2)]#