Solve equation with complex number

Question

How would you solve for x: $$ 2 = |x - 1 + 2i|$$ given that x $\geq$ 0

One way I guess can be, $$ 2 = \sqrt {(x-1)^2 + (2i)^2} \implies 4 = {(x-1)^2 + (2i)^2}$$

or can I split the absolute operator like this? $$ 2 = |x| + |-1 + 2i| $$

Answer 1

Looking geometrically, the complex points $x$ such that $2=|x-1+2i|=|x-(1-2i)|$ are those on the circle of radius $2$ with the centre in $1-2i$. If we are looking for a real $x$: this circle touches the $x$-axis at $x=1$ and so $x=1$ is the unique solution.

The algebraic solution also looks promising, you just needed to correct a few bits: $2=\sqrt{(x-1)^2+2^2}$, i.e. $4=(x-1)^2+4$, i.e. $(x-1)^2=0$ i.e. $x=1$.

Answer 2

$|x -1 + 2i| \ne \sqrt {(x-1) + (2i)^2}$.

You are simply wrong about that.

$|x - 1 + 2i| = \sqrt{(x - 1 + 2i)(\overline{x - 1+2i})} =$

$\sqrt{(Re(x-1+2i)+Im(x-1+2i)i)(Re(x-1+2i)-Im(x-1+2i))}$

$ =\sqrt{Re(x-1 +2i)^2 + Im(x-1+2i)^2}$

$= \sqrt{Re(x-1)^2 + [Im(x-1) + 2]^2}$.

But there is one HUGE hint. $x \ge 0$ and there is not order on complex numbers so $x \in \mathbb R$.

So $|x-1 + 2i| = \sqrt{(x-1 +2i)(x - 1 - 2i)} = \sqrt{(x-1)^2 + 2^2} = 2$

So $(x-1)^2 + 4 = 4$

So $(x-1)^2 = 0$.

So take it from there....

Answer 3

You need to remember that $$|a+bi|=\sqrt{a^2+b^2}$$ otherwise you would have strange results like $|1+i|=\sqrt{1-1}=0$, which doesn't make sense.

Knowing this, it's a little bit easier: $$2=|x-1+2i|=\sqrt{(x-1)^2+2^2}$$ Squaring both sides we get: $$4=(x-1)^2+4$$ $$(x-1)^2=0$$ So finally we get: $$x=1$$