Question 1083295
{{{r=sqrt((1)^2+(-3)^2)=sqrt(1+9)=sqrt(10)}}}
and
{{{theta=tan^(-1)(-3/1)=tan^(-1)(-3))=288.4}}}
So then,
{{{sqrt(1-3i)=sqrt(sqrt(10))(cos(288.4/2)+i*sin(288.4/2))}}}
{{{sqrt(1-3i)=sqrt(sqrt(10))(cos(144.2)+i*sin(144.2))}}}
{{{Z[1]=-1.442+1.040i}}}
and
{{{sqrt(1-3i)=sqrt(sqrt(10))(cos((288.4+360)/2)+i*sin((288.4+360)/2))}}}
{{{sqrt(1-3i)=sqrt(sqrt(10))(cos((324.2)/2)+i*sin((324.2)))}}}
{{{Z[2]=1.442-1.040i}}}