Q.Show that the equation $z^3+2\overline z=0$ has five solutions.

Writing $z$ in polar form, the imaginary part of $z+1/z$ must vanish; since $z$ is non-real this forces $r=1/r$, i.e. $|z|=1$.

1. Let $z=r(\cos\theta+i\sin\theta)$ with $r=|z|>0$. Since $z\in C\setminus R$, $\sin\theta\ne0$.
2. $\dfrac1z=\dfrac1r(\cos\theta-i\sin\theta)$ (since $\dfrac1{\cos\theta+i\sin\theta}=\cos\theta-i\sin\theta$).
3. $z+\dfrac1z=\left(r+\dfrac1r\right)\cos\theta+i\left(r-\dfrac1r\right)\sin\theta$. …

Rewrites each conjugate as $\bar z_k=|z_k|^2/z_k$ to recognise the given expression as $z_1z_2z_3(\bar z_1+\bar z_2+\bar z_3)$, then uses $|\bar w|=|w|$.

1. Since $|z_1|=1$: $\bar z_1=\dfrac{1}{z_1}$. Since $|z_2|=2$: $\bar z_2=\dfrac{4}{z_2}$. Since $|z_3|=3$: $\bar z_3=\dfrac{9}{z_3}$.
2. Consider $z_1z_2z_3(\bar z_1+\bar z_2+\bar z_3)=z_1z_2z_3\left(\dfrac1{z_1}+\dfrac4{z_2}+\dfrac9{z_3}\right)=z_2z_3+4z_1z_3+9z_1z_2$, which is exactly the given expression $9z_1z_2+4z_1z_3+z_2z_3$.
3. So $9z_1z_2+4z_1z_3+z_2z_3=z_1z_2z_3(\bar z_1+\bar z_2+\bar z_3)$, hence $|9z_1z_2+4z_1z_3+z_2z_3|=|z_1z_2z_3|\cdot|\bar z_1+\bar z_2+\bar z_3|$. …

Uses multiplicativity of the complex modulus on both sides of the given product, then squares to match $x^2+y^2$.

1. Given $(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy$.
2. Take modulus of both sides: $|1+i|\,|1+2i|\cdots|1+ni|=|x+iy|$.
3. Square both sides: $|1+i|^2|1+2i|^2\cdots|1+ni|^2=x^2+y^2$. …

Using $|z|=1\Rightarrow\bar z=1/z$ turns the given ratio into a simple algebraic simplification.

1. Since $|z|=1$, we have $z\bar z=|z|^2=1$, so $\bar z=\dfrac1z$. …

Taking the modulus-squared of $(1+i)(1+2i)\cdots(1+ni)=x+iy$ turns the product of $(1+k^2)$ terms into $x^2+y^2$.

1. We are given $(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy$.
2. Taking the modulus on both sides: $|1+i|\,|1+2i|\,|1+3i|\cdots|1+ni|=|x+iy|$.
3. Since $|1+ki|=\sqrt{1+k^2}$, the left side is $\sqrt{1+1^2}\sqrt{1+2^2}\sqrt{1+3^2}\cdots\sqrt{1+n^2}=\sqrt{2\cdot5\cdot10\cdots(1+n^2)}$. …

Comparing real and imaginary parts after conjugating gives $m=-8$ and $n=-\dfrac12$, i.e. $(n,m)=\left(-\dfrac12,-8\right)$.

1. The complex conjugate of $(2m+3)+i(3n-2)$ is $(2m+3)-i(3n-2)$.
2. Given: $(m-5)+i(n+4) = (2m+3)-i(3n-2)$.
3. Equate real parts: $m-5 = 2m+3 \Rightarrow -m = 8 \Rightarrow m=-8$. …