Question 1177967
{{{(-1-i*sqrt(3))^2}}}

{{{1+ 2sqrt(3)*i+i^2*3}}}

{{{1+ 2sqrt(3)*i-3}}}

{{{-2 + i 2sqrt(3) }}}


=> {{{z=a+ bi   }}} In your case


{{{a=-2 }}}and {{{b=2sqrt(3) }}}=> {{{a<0 }}} and {{{b>0}}}


To find argument  we use  the following formula:


{{{theta=tan^-1(b/a) +180}}}° if {{{ a<0}}}


{{{theta=tan^-1(2sqrt(3)/-2) +180}}}°


{{{theta=tan^-1(-sqrt(3)) +180}}}°


{{{theta=-60+180}}}°


 {{{theta=120}}}°