Express #z_1=sqrt(​3)-i# in polar form?

1 Answer
Shiva Prakash M V
May 10, 2018

#z_1=1angle(-pi/6)#
fourth quadrant

Explanation:

Given:
#z_1=sqrt3-i#
If #z=a+ib,#
#r=sqrt(a^2+b^2)#
and
#theta=tan^-1(b/a)#
Here,
#z=z_1, a=sqrt3, b=-1#
Thus,
#|z_1|=sqrt(sqrt3^2+(-1)^2#
#=sqrt(3+1)=sqrt4=2#
#theta_1=tan^-1((-1)/sqrt3)=-pi/6#

Hence, #z=r angletheta#

#z_1=1angle(-pi/6)#
fourth quadrant

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