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`
2. `A=5+6i,B=-2+3i,C=1-3i`
Find sqrt(A)Solution:Here `A=5+6i,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=5` and `b=6`
`:. r=sqrt(5^2+6^2)=sqrt(25+36)=sqrt(61)=7.8102`
`theta=atan(b/a)` (Since `a>0`)
`:. theta=atan((6)/(5))`
`:. theta=atan(1.2)`
`:. theta=50.1944 ^circ` or `theta=0.8761` rad
`:. theta=0.8761`
Exponential form:`5+6i=r*e^(i*theta)`
`5+6i=7.8102*e^(i(0.8761))`
Step-2: Apply the square root formulaNow `(5+6i)^(1/2)=(7.8102)^(1/2)*e^(i(1/2*0.8761))`
`=2.7947*e^(i(0.438))`
Step-3: Convert back to rectangular form`=2.7947*(cos(0.438)+isin(0.438))`
`=2.7947*(0.9056+0.4242i)`
`=2.5308+1.1854i`
3. `A=5+6i,B=-2+3i,C=1-3i`
Find sqrt(B)Solution:Here `A=5+6i,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=-2` and `b=3`
`:. r=sqrt((-2)^2+3^2)=sqrt(4+9)=sqrt(13)=3.6056`
`theta=atan(b/a)+180` (Since `a<0`)
`:. theta=atan((3)/(-2))+180`
`:. theta=atan(-1.5)+180`
`:. theta=-56.3099+180`
`:. theta=123.6901 ^circ` or `theta=2.1588` rad
`:. theta=2.1588`
Exponential form:`-2+3i=r*e^(i*theta)`
`-2+3i=3.6056*e^(i(2.1588))`
Step-2: Apply the square root formulaNow `(-2+3i)^(1/2)=(3.6056)^(1/2)*e^(i(1/2*2.1588))`
`=1.8988*e^(i(1.0794))`
Step-3: Convert back to rectangular form`=1.8988*(cos(1.0794)+isin(1.0794))`
`=1.8988*(0.4719+0.8817i)`
`=0.896+1.6741i`
This material is intended as a summary. Use your textbook for detail explanation.
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