Find such a complex number $z$ so that $|z| - 2 = z +12i$

Question

I have no idea how to approach this exercise. I've tried deducing the real component $a$ and imaginary component $bi$ by inserting $z = a + bi$ and $|z| = \sqrt{a^2 +b^2}$ into the original equation, but it just gets way longer and absurdly complicated. Nothing cancels out. This can't be the right strategy. Help please?

Answer 1

Hint. Let $z=x+iy$. Then, after separating the real and the complex parts, the equation becomes $$(-\sqrt{x^2+y^2}+2+x)+i(y+12)=0\Leftrightarrow \begin{cases}y+12=0\\ -\sqrt{x^2+y^2}+2+x=0 \end{cases}$$ Hence from the first equation we find $y=-12$. Now plug it into the second one and find $x$.

Answer 2

Let $$z=x+iy$$ then we have to solve $$\sqrt{x^2+y^2}-2=x+i(y+12)$$

Answer 3

$$|Z| - 2 = Z +12i \implies Z=a-12i $$

$$ |Z| = \sqrt {a^2 +144} =Z+12i+2 = a+2$$

$$ a^2+144 = a^2+4a+4$$

$$a=35$$

$$Z=35-12i$$

Answer 4

As an alternative by conjugate, note that

$$|z|-2=z+12i\iff |z|-2=\bar z-12i$$

and subtracting

$$z-\bar z=-24i\iff \frac{z-\bar z}{2i}=-12 \implies y=-12$$

and adding

$$z+\bar z=2|z|-4 \\\implies 2x=2\sqrt{x^2+144}-4\iff x+2=\sqrt{x^2+144}\\\iff 4x=140\implies x=35 $$

Answer 5

Rearranging the equation and taking conjugates on both sides:

$$ \begin{align} z = |z| - 2 - 12i \tag{1}\\ \bar z = |z| - 2 + 12i \end{align} $$

Multiplying the above:

$$ |z|^2=\big(|z|-2\big)^2 + 144 \;\;\iff\;\; 4 |z| - 148 = 0 \;\;\iff\;\; |z| = 37 \tag{2} $$

Replacing $(2)$ back in $(1)$ gives $\,z=37-2-121i=35-12i\,$.