How do you write the complex number in trigonometric form #5+2i#?

Question

↳Redirected from "What is Boyle's law?"

4279 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-19T16:54:43

#5+2i=sqrt29costheta+isqrt29sintheta#, where #theta=tan^(-1)(2/5)#

A number #a+ib# can be written in trigonometric form as

#rcostheta+irsintheta#.

As #rcostheta=a# and #rsintheta=b#, squaring and adding them we get #r^2=a^2+b^2#.

As such for #5+2i#, #r=sqrt(5^2+2^2)=sqrt(25+4)=sqrt29#

and #costheta=5/sqrt29# and #sintheta=2/sqrt29#

i.e. #tantheta=2/5# and #theta=tan^(-1)(2/5)#

Hence in trigonometric form

#5+2i=sqrt29costheta+isqrt29sintheta#, where #theta=tan^(-1)(2/5)#