It depends on what you get if are ranked low you will need help with the skill and if rank average you just are in normal classes and if you are ranked above average you probably are either left in the average class or go to advance math.

6 0

3 years ago

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

leonid [27]

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

4 0

3 years ago

Pls answer this there's a pic above​

Pls answer this there's a pic above