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

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?

• Hint: $|z|-2$ is real – BAI Apr 14 '18 at 13:29

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 $x$.

• Aaaaah, yes, I've forgotten about this essential property of the complex numbers. Something along the lines of Vieta's Formula, but better. Thanks! I think I'll manage from here. Just one more thing: the main idea is to solve for $y$ and the insert the deduced value into second equation, right? – Gregor Perčič Apr 14 '18 at 13:26
• Yes, that's correct. – Robert Z Apr 14 '18 at 13:27

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

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

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

$$a=35$$

$$Z=35-12i$$

As an alternative by conjugate, note that

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

and subtracting

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

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

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 $\,z=37-2-121i=35-12i\,$.

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