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17. Powers of complex numbers example ( Enter your problem )

1. Example-1





1. `A=5+6i`
Find pow(A,6)


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)^(6)=(7.8102)^(6)*e^(i(6*0.8761))`

`=226981*e^(i(5.2563))`

Step-3: Convert back to rectangular form
`=226981*(cos(5.2563)+isin(5.2563))`

`=226981*(0.5175-0.8557i)`

`=117469-194220i`
2. `A=-2+3i`
Find pow(A,4)


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)^(4)=(3.6056)^(4)*e^(i(4*2.1588))`

`=169*e^(i(8.6352))`

Step-3: Convert back to rectangular form
`=169*(cos(8.6352)+isin(8.6352))`

`=169*(-0.7041+0.7101i)`

`=-119+120i`
3. `A=1-3i`
Find pow(A,3)


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

`=31.6228*e^(i(-3.7471))`

Step-3: Convert back to rectangular form
`=31.6228*(cos(-3.7471)+isin(-3.7471))`

`=31.6228*(-0.8222+0.5692i)`

`=-26+18i`




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