Question 1180603

Given:

{{{z[1]=sqrt(3)-i}}}

If  {{{z=a+b*i}}}, then

{{{r=sqrt(a^2+b^2)}}}

and

{{{theta=tan^-1(b/a)}}}

Here,

{{{z=z[1]}}},{{{a=sqrt(3)}}}, and {{{b=-1}}}


Thus,

{{{r=sqrt((sqrt(3))^2+(-1)^2)}}}

{{{r=sqrt(3+1)}}}

{{{r=sqrt(4)}}}

{{{r=2}}}


{{{theta=tan^-1(-1/sqrt(3))}}}

{{{theta=-pi/6}}}

Hence, 


The polar form is ({{{2}}},{{{-pi/6}}})