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16. Cube root of complex number example ( Enter your problem )

1. Example-1





1. `A=5+6i`
Find Cube root of complex number


Solution:
Here `A=5+6i`

For a complex number `z=a+bi`, the polar form is `z=r*(cos(theta)+i*sin(theta))`

then power of n of given complex number can be obtained by
`z^n=[r*(cos(theta)+i*sin(theta))]^n=r^n*[cos(n*theta)+i*sin(n*theta)]`


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 power formula
Now `(5+6i)^(0.3333)=(7.8102)^(0.3333)*e^(i(0.3333*0.8761))`

`=1.9841*e^(i(0.292))`

Step-3: Convert back to rectangular form
`=1.9841*(cos(0.292)+isin(0.292))`

`=1.9841*(0.9577+0.2879i)`

`=1.9001+0.5712i`
2. `A=-2+3i`
Find Cube root of complex number


Solution:
Here `A=-2+3i`

For a complex number `z=a+bi`, the polar form is `z=r*(cos(theta)+i*sin(theta))`

then power of n of given complex number can be obtained by
`z^n=[r*(cos(theta)+i*sin(theta))]^n=r^n*[cos(n*theta)+i*sin(n*theta)]`


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 power formula
Now `(-2+3i)^(0.3333)=(3.6056)^(0.3333)*e^(i(0.3333*2.1588))`

`=1.5334*e^(i(0.7196))`

Step-3: Convert back to rectangular form
`=1.5334*(cos(0.7196)+isin(0.7196))`

`=1.5334*(0.7521+0.6591i)`

`=1.1532+1.0106i`
3. `A=1-3i`
Find Cube root of complex number


Solution:
Here `A=1-3i`

For a complex number `z=a+bi`, the polar form is `z=r*(cos(theta)+i*sin(theta))`

then power of n of given complex number can be obtained by
`z^n=[r*(cos(theta)+i*sin(theta))]^n=r^n*[cos(n*theta)+i*sin(n*theta)]`


Step-1: Convert to exponential form: `z = re^(i theta)`

Here, `a=1` and `b=-3`

`:. r=sqrt(1^2+(-3)^2)=sqrt(1+9)=sqrt(10)=3.1623`

`theta=atan(b/a)` (Since `a>0`)

`:. theta=atan((-3)/(1))`

`:. theta=atan(-3)`

`:. theta=-71.5651 ^circ` or `theta=-1.249` rad

`:. theta=-1.249`

Exponential form:
`1-3i=r*e^(i*theta)`

`1-3i=3.1623*e^(i(-1.249))`

Step-2: Apply the power formula
Now `(1-3i)^(0.3333)=(3.1623)^(0.3333)*e^(i(0.3333*-1.249))`

`=1.4678*e^(i(-0.4163))`

Step-3: Convert back to rectangular form
`=1.4678*(cos(-0.4163)+isin(-0.4163))`

`=1.4678*(0.9146-0.4044i)`

`=1.3424-0.5936i`




This material is intended as a summary. Use your textbook for detail explanation.
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