Question

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

Answer 1

Answer:

$b = 3$

Step-by-step 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.