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Question

There is a complex number $z$ with imaginary part $164$ and a positive integer n such that $\frac{z}{z + n} = 4 i .$ The value of $n$ is

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Solution

$\frac{z}{z + n} = 4 i$
Let $z = x + i y \Rightarrow x + 164 i$
$\frac{x + 164 i}{x + 164 i + n} = 4 i$
$x + 164 i = 4 i \times (x + n + 164 i)$ $i^{2} = - 1$
$x + 164 i = (4 (x + n) i - 656)$
on comparing
x = -656
$4 (x + n) = 164$
$(n - 656) = 41$
n = 656 + 41 = 697

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