Home > Calculus calculators > Cube root of complex number example

15. Cube root of complex number example ( Enter your problem )
  1. Example-1
Other related methods
  1. Adding complex numbers
  2. Subtracting complex numbers
  3. Multiplying complex numbers
  4. Dividing complex numbers
  5. Real part of complex number
  6. Imaginary part of complex number
  7. Conjugate of complex number
  8. Modulus / Magnitude / Absolute value of complex numbers
  9. Multiplicative inverse of complex numbers
  10. Reciprocal of complex number
  11. Argument / Arg / Angle / Phase of complex number
  12. Polar form of complex numbers
  13. Exponential form of complex numbers
  14. Square root of complex number
  15. Cube root of complex number
  16. Powers of complex numbers
  17. Roots of complex numbers

14. Square root of complex number
(Previous method)
16. Powers of complex numbers
(Next method)

1. Example-1





1. `A=-1-i,B=-2+3i,C=1-3i`
Find cbrt(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 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=-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=1.25pi` 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 power formula
Now `(-1-i)^(0.3333)=(1.4142)^(0.3333)*e^(i(0.3333*3.927))`

`=1.1225*e^(i(1.309))`

Step-3: Convert back to rectangular form
`=1.1225*(cos(1.309)+isin(1.309))`

`=1.1225*(0.2588+0.9659i)`

`=0.2905+1.0842i`
2. `A=5+6i,B=-2+3i,C=1-3i`
Find cbrt(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 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`
3. `A=5+6i,B=-2+3i,C=1-3i`
Find cbrt(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 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`


This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here



14. Square root of complex number
(Previous method)
16. Powers of complex numbers
(Next method)





Share this solution or page with your friends.
 
 
Copyright © 2025. All rights reserved. Terms, Privacy
 
 

.