Question

The product 2+bi)(3+2i) is equal to 13i. What is the value b?

Answer 1

`b = 3`

Explanation:

We are given that:

`(2 + bi)(3+2i) = 13i`

And we want to determine the value of b.

First, expand:

`\displaystyle \begin{aligned} (2+bi)(3+2i)&=2(3+2i)+bi(3+2i) \\ &= (6+4i)+(3bi+2bi^2) \\ &= (6+2b(-1))+(4i+3bi) \\ &= (6-2b) + (4+3b)i\end{aligned}`

Therefore, we can write that:

`(6-2b) + (4+3b)i = 0+ 13i`

If two complex numbers are equivalent, their real and imaginary parts must be equivalent. Hence:

`6-2b = 0 \text{ and } 4+3b = 13`

Solve for each case:

`b=3 \text{ and } b=3`

The two solutions are equivalent, hence such a number b exists.

In conclusion, b = 3.