Question

Let $a$ and $b$ be nonzero real numbers such that

Answer 1

Answer:   -7/2

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Explanation:

Let's expand out (2 - 7i)(a + bi) using the FOIL rule

(2 - 7i)(a + bi) = 2a + 2bi - 7ai - 7bi^2

(2 - 7i)(a + bi) = 2a + 2bi - 7ai - 7b(-1)

(2 - 7i)(a + bi) = 2a + 2bi - 7ai + 7b

(2 - 7i)(a + bi) = (2a+7b) + (2bi-7ai)

(2 - 7i)(a + bi) = (2a+7b) + (2b-7a)i

We're told the result is purely imaginary. What this means is that the real part (2a+7b) is zero, while the imaginary part (2b-7a) is nonzero. If both are zero, then we have 0+0i = 0 which is purely real.

For example, the complex numbers 0-7i and 0+2i are purely imaginary.

Let's use the fact that 2a+7b must be zero to do the following steps:

2a+7b = 0

2a = -7b

a = -7b/2

a/b = -7/2 which is the final answer

We must check to see if 2b-7a is nonzero

2b - 7a = 2b - 7(-7b/2)

2b - 7a = 2b + 24.5b

2b - 7a = 26.5b

The result is nonzero if and only if b is nonzero. Luckily we're told b is nonzero at the top of the problem. So we don't have any worries that (2b-7a) is zero.

Therefore, (2a+7b) + (2b-7a)i will be purely imaginary with a/b = -7/2

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A concrete example:

Let a = -14 and b = 4

a/b = -14/4 = -7/2

(2-7i)(a+bi) = (2-7i)(-14+4i) = 0 + 106i which is purely imaginary.

I'll let you do the steps in expanding that out using the FOIL rule.