Question

Find the multiplicative inverse of 3 − 2i. Verify that your solution is corect by confirming that the product of

Answer 1

Answer:

$\frac{3}{13} + \frac{2i}{13}$

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 = $\frac{1}{3-2i}$

We have that

$(3- 2i)\frac{1}{3-2i} = 1$ would be the multiplicative inverse of 3 - 2i

But remember that 2i = √-2 so we can rationalize the denominator of this complex number

$\frac{1}{3-2i } (\frac{3+2i}{3+2i } )=\frac{3+2i}{9-(4i^{2} )} =\frac{3+2i}{9-4(-1)} =\frac{3+2i}{13}$

Thus, the multiplicative inverse would be $\frac{3}{13} + \frac{2i}{13}$

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:

$(3-2i)(\frac{3+2i}{13}) = \frac{9-4i^{2} }{13} =\frac{9-4(-1)}{13}= \frac{9+4}{13} = \frac{13}{13}= 1$

Thus, the number we found is indeed the multiplicative inverse of  3 - 2i