# The multiplicative inverse of a complex number is -2x + 3i, find the complex number and its additive inverse.show complete solution and explain the answer.

Asked on by spock7

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The multiplicative inverse of a complex number is -2x + 3i. If the complex number is N, N*(-2x + 3i) = 1

=> N = 1/(-2x + 3i)

=> N = (-2x - 3i)/(-2x + 3i)(-2x - 3i)

=> N = (-2x - 3i)/((-2x)^2 - (3i)^2)

=> N = (-2x - 3i)/(4x^2 + 9)

The additive inverse is the number that gives 0 when it is added to the number, or 0 - (-2x - 3i)/(4x^2 + 9)

=> (2x + 3i)/(4x^2 + 9)

The complex number is (-2x - 3i)/(4x^2 + 9) and the additive inverse is (2x + 3i)/(4x^2 + 9)

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You know that the multiplicative inverse of a complex number z is 1/z.

Therefore -2x+3i=1/z =>z=1/(-2x+3i)

You must not keep a complex number to denominator so you must multiply the fraction 1/(-2x+3i) by the conjugate -2x-3i.

z=(-2x-3i)/(-2x+3i)(-2x-3i)

z=(-2x-3i)/(4x^2 + 9)

The complex number is z=-2x/(4x^2 + 9)-i*3/(4x^2 + 9)

The additive inverse of a complex number z is denoted by -z so that:

z+(-z)=0

The additive inverse of z is: -z=-[2x/(4x^2 + 9)-i*3/(4x^2 + 9)]

-z=2x/(4x^2 + 9)+i*3/(4x^2 + 9)

Answer: The complex number is z= -2x/(4x^2 + 9)-i*3/(4x^2 + 9) and the additive inverse is -z=2x/(4x^2 + 9)+i*3/(4x^2 + 9).

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