# Characterize the set of complex $z$ such that $a + \frac{a+8b}{a^2 + b^2} = 2$ and $b + \frac{8a-b}{a^2 + b^2} = 0$

Characterize the set of $$z = a + bi$$ such that $$a + \frac{a+8b}{a^2 + b^2} = 2\\ b + \frac{8a-b}{a^2 + b^2} = 0$$ Source: Art of Problem Solving

I've made progress but been unable to complete this problem correctly. Can you help me figure out my mistake or take a better approach?

In general, the approach I've taken has proven difficult, because the manipulations grow too complex to do correctly (even after many rounds of checking), but I can find no approach that avoids them.

Additionally: I normally try to have some qualitative description of what the equations tell me or where I'm going, but am unable to do so here. The only way I saw to proceed here was to replace $$a$$ and $$b$$ with their $$z, \bar z$$ extractions, so that everything is in the same single variable $$z$$. Is there an insight, a picture, a realization relevant here that can replace blind manipulations?

My work so far: Let $$w = \frac {1 + 8i}{z}, v = \frac {1 - 8i}{z}$$. Then since $$a = \frac{z + \bar z }{2}, b= \frac{z - \bar z}{2i}$$, we have $$a + \frac{a+8b}{a^2 + b^2} = 2 \\ z + \bar z + \frac{z + \bar z - 8i(z - \bar z)}{z \bar z} = 4\\ z + \bar z + w + \bar w = 4$$ and $$b + \frac{8a-b}{a^2 + b^2} = 0\\ -i(z - \bar z) + \frac{8(z + \bar z) + i(z - \bar z) }{z \bar z} = 0 \\ % -i[z - \bar z + \frac{-8zi - 8 \bar z i - z + \bar z}{z \bar z}] = 0 \\ % z - \bar z + \frac{\bar z - 8 \bar z i - (z + 8zi)}{z \bar z} = 0 \\ z - \bar z + \frac {1 - 8i}{z} - \frac {1 + 8i}{\bar z}= 0\\ z - \bar z + v - \bar v= 0$$ so $$2z + w + v + \bar w - \bar v = 4\\ z + \frac 1 z = 2 \\ z^2 - 2z + 1 = 0\\ (z-1)(z - 1) = 0 \\ z = 1.$$

However, that does not meet the original equations.

What is my error? Is there a better approach?

## Update

With help from "SwagBeastSKJ", this should be: $$2z + w + v + \bar w - \bar v = 4\\ z + \frac 1 z = 2 - 4i\\ z^2 + (-2 + 4i)z + 1 = 0$$

I could then use the quadratic formula... but something tells me I'm missing the boat. There must be an elegant approach, an insight, that I'm missing.

• Why not discuss this at Art of Problem Solving ?? Jul 14, 2023 at 0:54
• How did you go from $2z + w + v + \bar{w} - \bar{v}$ to $z + \frac{1}{z}$ Jul 14, 2023 at 0:55
• @SwagBeastSKJJ Thank you. I thought $w + v = \frac 2 z, \bar w - \bar v = 0$. Then divide both sides by 2. But I see $\bar w - \bar v = -8i$. Let me work this... Jul 14, 2023 at 1:00
• I think you have $\bar{w} - \bar{v}$ wrong. Jul 14, 2023 at 1:02

We have $$z = a + bi$$

$$a + \frac{a + 8b}{a^2 + b^2} = 2$$ $$b + \frac{8a - b}{a^2 + b^2} = 0$$

Multiplying second equation by $$i^3$$ we get

$$-bi + \frac{-8ai + bi}{a^2 + b^2} = 0$$

Adding with our first equation gives us

$$a - bi + \frac{a - 8ai + 8b + bi}{a^2 + b^2} = 2$$

$$\implies \bar{z} + \frac{z - 8zi}{z\bar{z}} = 2$$

$$\implies \bar{z} + \frac{1-8i}{\bar{z}} = 2$$

Which is a quadratic equation with complex roots in $$\bar{z}$$ and can be easily solved.

This gives us solutions $$\bar{z} = -1 - 2i$$ and $$\bar{z} = 3 + 2i$$ which means $$z = -1 + 2i$$ and $$z = 3 - 2i$$, and checking the solutions we find this is indeed the solution to the given equation.