# What is the multiplicative inverse of the number 3 + 2i ?

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The multiplicative inverse of 3 + 2i is the number that if multiplied by 3 + 2i gives 1.

If the multiplicative inverse is x + iy.

(x + iy)(3 + 2i) = 1

=> x +iy = 1/(3 +2i)

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

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

=> x + iy = (3 - 2i)/ 13

We get x + iy = 3/13 - 2i/13

**The multiplicative inverse is 3/13 - 2i/13**

We'll obtain the multiplicative inverse when multiplying the given complex number by the inverse number we'll get the result 1.

We'll note the inverse as x:

(3 + 2i)*x = 1

We'll divide by (3 + 2i) both sides:

x = 1/(3 + 2i)

Since, we are not allowed to keep a complex number to denominator, we'll multiply the ratio by the conjugate of the complex number:

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

We'll re-write the denominator as a difference of squares:

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

(3 + 2i)(3 - 2i) = 9 - 4i^2

But i^2 = -1:

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

(3 + 2i)(3 - 2i) = 13

We'll re-write x:

x = (3 - 2i)/13

x = 3/13 - 2i/13

**The multiplicative inverse of the complex number 3 + 2i is 3/13 - 2i/13.**