# How do you add #(2-3i)# and #(12-2i)# in trigonometric form?

##### 1 Answer

May 17, 2016

#### Explanation:

A complex number z = x +iy can be expressed in trig. form as shown.

#z=x+iy=r(costheta+isintheta)" where"#

#•r=sqrt(x^2+y^2)" and " theta=tan^-1(y/x)# Now to get this sum in trig form we have to add the numbers together and then convert to trig.

#rArr(2-3i)+(12-2i)=14-5i# Using x = 14 and y = -5 , convert to trig form.

#rArrr=sqrt(14^2+(-5)^2)=sqrt221" does not simplify further"# and

#theta=tan^-1(-5/14)≈-0.343" radians"#

#rArr14-5i=sqrt221(cos(-0.343)+isin(-0.343))# using

#cos(-theta)=costheta" and "sin(-theta)=-sintheta# we can also express in trig form as

#14-5i=sqrt221(cos(0.343)-isin(0.343))#