Squaring complex number equation with absolute values

Question

I don't understand how you go from the first to the second line in this problem :

$$|(a-k)+i(7-2a)|=|(a-2)+i(9-2a)|$$

$$(a-k)^2+(7-2a)^2=(a-2)^2+(9-2a)^2.$$

Firstly, squaring i should make it -1 I believe, it just seemed to disappear here? Second, shouldn't the square be applied to the whole of the left and right parts of the equation and not just the individual parts between brackets?

Answer 1

In addition to what @Kavi Rama Murthy said and noting that in both the RHS and LHS, the expressions in the absolute value function are just two complex numbers; $a$ and $k$ are just constants, recall the definition of the absolute value in the argand diagram.

• $z$ is the net vector addition of $x$ and $y$, hence $z = x+iy$
• The distance or magnitude of the vector $z$ is the same as the $|z|$