1. `A=-1-i,B=-2+3i,C=1-3i`
Find sqrt(A)Solution:Here `A=-1-i,B=-2+3i,C=1-3i`
For a complex number `z=a+bi`, the polar form is `z=r*(cos(theta)+i*sin(theta))`
then square roots of given complex number can be obtained by
`z^(1/2)=[r*(cos(theta)+i*sin(theta))]^(1/2)=sqrt(r)*[cos((theta)/2)+i*sin((theta)/2)]`
Step-1: Convert to exponential form: `z = re^(i theta)`Here, `a=-1` and `b=-1`
`:. r=sqrt((-1)^2+(-1)^2)=sqrt(1+1)=sqrt(2)=1.4142`
`theta=atan(b/a)+180` (Since `a<0`)
`:. theta=atan((-1)/(-1))+180`
`:. theta=atan(1)+180`
`:. theta=45+180`
`:. theta=225 ^circ` or `theta=(5pi)/(4)` rad = 3.927 rad
`:. theta=3.927`
Exponential form:`-1-i=r*e^(i*theta)`
`-1-i=1.4142*e^(i(3.927))`
Step-2: Apply the square root formulaNow `(-1-i)^(1/2)=(1.4142)^(1/2)*e^(i(1/2*3.927))`
`=1.1892*e^(i(1.9635))`
Step-3: Convert back to rectangular form`=1.1892*(cos(1.9635)+isin(1.9635))`
`=1.1892*(-0.3827+0.9239i)`
`=-0.4551+1.0987i`
