The multiplicative inverse of a number X is a number Y such that X*Y=1.

Here X= (7+2i)/(6-3i)

Therefore the multiplicative inverse is (6-3i)/(7+2i).

(6-3i)/(7+2i)= [(6-3i)*(7-2i)]/[(7-2i)/(7+2i)]

= [42-12i-21i+6i^2]/[7^2-4*i^2]

=[42-33i-6]/[49+4]

=(36-33i)/53

**The required multiplicative inverse is (36-33i)/53**

We'll establish the multiplicative inverse of the given ratio is:

(6-3i)/(7+2i)

According to the rule, it is not allowed that the denominator to be a complex number.

We'll multiply the ratio by the conjugate of (7+2i).

The conjugate of 7+2i = 7-2i

(6-3i)/(7+2i) = (6-3i)(7-2i) / (7+2i)(7-2i)

We'll remove the brackets:

(6-3i)(7-2i) = 42 - 12i - 21i - 6

We'll combine the real parts and the imaginary parts and we'll get:

Now, we'll calculate the difference of squares:

(7+2i)(7-2i) = (7)^2 - (4i)^2 = 49 + 16 = 65

**(6-3i)/(7+2i) = ( 36 - 33i)/65**

**The multiplicative inverse of the number (7+2i)/(6-3i) is:**

**36/65 - (33/65)*i**

The multiplicative inverse of a number X is equal to `1/X` .

For `X = (7+2i)/(6-3i)` , the multiplicative inverse is `1/X = 1/((7+2i)/(6-3i))`

= `(6-3i)/(7+2i)`

To simplify this multiply the numerator and denominator with the complex conjugate of the denominator.

`((6-3i)*(7-2i))/((7+2i)*(7-2i))`

= `((6-3i)*(7-2i))/(7^2 + 2^2)`

= `(42 - 21i - 12i + 6i^2)/(53)`

= `(42 - 21i - 12i - 6)/(53)`

= `(36 - 33i)/(53)`

To find the multiplicative inverse of (7+2i)(6-3i).

The multiplicative inverse of z = (a+bi) is z' and zz' = 1.

Or z' = 1/z.

First we simplify (7+2i)(6-3i) = 42 -21i+12i -6i^2

= 42-9i +6, as i^2 = -1

=48 -9i

Therefore z = (48-9i

z' = 1/(48-9i). Realise the denomonator by multiplying (48+9i) both numerator and denominators.

z' = (48+9i)/(48^2+9^2)

z' = (48/2385) - (9/2385)i

z' = 16/795 -(3/795)i