Find all complex numbers $z$ such that $\frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}$ is a real number.

Question

Find all complex numbers $z$ such that:

$\frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}$

is a real number. I'm preparing for a math competition, but I can't solve this one. Can anyone help me? Thanks in advance. :)

Answer 1

For a complex number to be purely real, $z=\bar{z}$. So, taking $x=z^2,$ $$Y=\frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=\frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}$$ $$Y=\bar{Y}\implies \frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}=\frac{6\bar{x^2} + 5\bar{x} + 6}{3\bar{x^2} + 10\bar{x} + 3}$$ Cross multiply, and cancel lot of terms to get $$(x\bar{x}-1)(x-\bar{x})=0$$ $$x=\bar{x} \implies z=\pm \bar{z}$$ $$x\bar{x}=1\implies |z|=1.$$

Answer 2

Hint: Plug in your formula $$z=x+iy$$ then you will get$$24x^3y-24xy^3+10xy=0$$ and $$12x^3y-12xy^3+20xy=0$$

Answer 3

This is probably too much of a hint, but is based on my comment. I suggest dividing numerator and denominator by $z^2$ and setting $y=z^2+\frac 1{z^2}$ when your fraction becomes $$\frac {6y+5}{3y+10}$$ and it is easy now to show that $y$ has to be real, and if $y$ is real so is the fraction. So set $y=r$, an arbitrary real number, whence $z^4-rz^2+1=0$.

This can be solved in various ways, which include $$z^4-2z^2+2-(r-2)z^2=(z^2+z\sqrt {r-2}+1)(z^2-z\sqrt {r-2}+1)=0$$

If we now put $r-2=4R$ with $R$ a real number, this has solutions $$z=\pm \sqrt R\pm \sqrt {R-1}$$and though this is a clunky way through in the end, we get $z$ real, or $z$ pure imaginary, or $|z|=1$ when $R\in [0,1]$.

Others have found easier ways through to these conditions.

Answer 4

You can express: $$\frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=2-\frac{15z^2}{3z^4 + 10z^2 + 3}=2-\frac{15}{3z^2+10+\frac3{x^2}}=2-\frac{15}{3(z+\frac1z)^2+4} \Rightarrow \\ \left(z+\frac1z\right)^2=\left(a+bi+\frac1{a+bi}\right)^2=\left(a+bi+\frac{a-bi}{a^2+b^2}\right)^2 \Rightarrow \\ 1) \ a^3+ab^2+a=0 \ \ \text{or} \ \ 2) \ a^2b+b^3-b=0 \Rightarrow \\ 1) \ a=0 \ \ \text{or} \ \ 2) \ b=0 \ \text{or} \ a^2+b^2=1.$$