Answer:
[tex]\frac{3}{13} + \frac{2i}{13}[/tex]
Step-by-step explanation:
The multiplicative inverse of a complex number y is the complex number z such that (y)(z) = 1
So for this problem we need to find a number z such that
(3 - 2i) ( z ) = 1
If we take z = [tex]\frac{1}{3-2i}[/tex]
We have that
[tex](3- 2i)\frac{1}{3-2i} = 1[/tex] would be the multiplicative inverse of 3 - 2i
But remember that 2i = √-2 so we can rationalize the denominator of this complex number
[tex]\frac{1}{3-2i } (\frac{3+2i}{3+2i } )=\frac{3+2i}{9-(4i^{2} )} =\frac{3+2i}{9-4(-1)} =\frac{3+2i}{13}[/tex]
Thus, the multiplicative inverse would be [tex]\frac{3}{13} + \frac{2i}{13}[/tex]
The problem asks us to verify this by multiplying both numbers to see that the answer is 1:
Let's multiplicate this number by 3 - 2i to confirm:
[tex](3-2i)(\frac{3+2i}{13}) = \frac{9-4i^{2} }{13} =\frac{9-4(-1)}{13}= \frac{9+4}{13} = \frac{13}{13}= 1[/tex]
Thus, the number we found is indeed the multiplicative inverse of 3 - 2i
