How do you express #z = 1 + i sqrt(3)# in polar and trigonometric form, and sketch z on the complex number plane?

Question

32598 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-18T13:27:53

The polar form is #(2, pi/3)#. The trigonometric form is #z=2(cos(pi/3)+isin(pi/3))#

The polar form is #r=f(theta)#

#z=costheta+isintheta#

The modulus of #z# is

#|z|=|1+sqrt(3)i|=sqrt(1+(sqrt3)^2)=sqrt4=2#

#r=z/(|z|)=(2)(1/2+sqrt(3)/2i)#

So,

#costheta=1/2# and #sintheta=sqrt(3)/2#

Therefore,

#theta=pi/3# #mod 2pi#

So,

The polar form is #(2,pi/3)#

The trigonometric form is #z=2(cos(pi/3)+isin(pi/3))#

graph{((x-1)^2+(y-sqrt3)^2-0.001)=0 [-1.313, 4.163, -0.42, 2.316]}