Home > Calculus calculators > Multiplicative inverse of complex numbers example

9. Multiplicative inverse of complex numbers 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

8. Modulus / Magnitude / Absolute value of complex numbers
(Previous method)
10. Reciprocal of complex number
(Next method)

1. Example-1





1. `A=5+6i,B=-2+3i,C=1-3i`
Find inv(A)


Solution:
Here `A=5+6i,B=-2+3i,C=1-3i`


For a complex number `z=a+bi`, the inverse is `z^(-1)=(a-bi)/(a^2+b^2)`



`z=a+bi`

`:. z^(-1)=1/(a+bi)`

`:. z^(-1)=1/((a+bi))*((a-bi))/((a-bi))`

`:. z^(-1)=(a-bi)/(a^2+b^2)`


`A=5+6i`

`:. A^(-1)=1/(5+6i)`

`=1/((5+6i))*((5-6i))/((5-6i))`

`=(5-6i)/(5^2+6^2)`

`=(5-6i)/(25+36)`

`=(5-6i)/61`

`=0.082-0.0984i`

`:.` The inverse of `5+6i` is `0.082-0.0984i`
2. `A=5+6i,B=-2+3i,C=1-3i`
Find inv(B)


Solution:
Here `A=5+6i,B=-2+3i,C=1-3i`


For a complex number `z=a+bi`, the inverse is `z^(-1)=(a-bi)/(a^2+b^2)`



`z=a+bi`

`:. z^(-1)=1/(a+bi)`

`:. z^(-1)=1/((a+bi))*((a-bi))/((a-bi))`

`:. z^(-1)=(a-bi)/(a^2+b^2)`


`B=-2+3i`

`:. B^(-1)=1/(-2+3i)`

`=1/((-2+3i))*((-2-3i))/((-2-3i))`

`=(-2-3i)/((-2)^2+3^2)`

`=(-2-3i)/(4+9)`

`=(-2-3i)/13`

`=-0.1538-0.2308i`

`:.` The inverse of `-2+3i` is `-0.1538-0.2308i`
3. `A=5+6i,B=-2+3i,C=1-3i`
Find inv(C)


Solution:
Here `A=5+6i,B=-2+3i,C=1-3i`


For a complex number `z=a+bi`, the inverse is `z^(-1)=(a-bi)/(a^2+b^2)`



`z=a+bi`

`:. z^(-1)=1/(a+bi)`

`:. z^(-1)=1/((a+bi))*((a-bi))/((a-bi))`

`:. z^(-1)=(a-bi)/(a^2+b^2)`


`C=1-3i`

`:. C^(-1)=1/(1-3i)`

`=1/((1-3i))*((1+3i))/((1+3i))`

`=(1+3i)/(1^2+3^2)`

`=(1+3i)/(1+9)`

`=(1+3i)/10`

`=0.1+0.3i`

`:.` The inverse of `1-3i` is `0.1+0.3i`


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



8. Modulus / Magnitude / Absolute value of complex numbers
(Previous method)
10. Reciprocal of complex number
(Next method)





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

.