If $z$ is a root of the equation $\dfrac{2z+1}{\bar{z}+i}=i$, find $|z|$.
I could solve this problem, but I'm curious to see if there are other methods to solve this problem. Here is my approach:
Let $z=x+yi$,
$$\dfrac{2z+1}{\bar{z}+i}=\dfrac{(2x+1)+2yi}{x+(-y+1)i}=i$$ $$(2x+1)+2yi=xi+(-y+1)i^2$$ $$2x+1=y-1 \qquad\text{And}\qquad 2y=x$$ Now substituting $x=2y$ in the first equation gives $4y+1=y-1$. Hence $y=-2/3$ and $x=-4/3$ which implies $|z|=\sqrt{20}/3=2\sqrt5/3$
We have $2z+1=i\bar z-1$. Conjugate to get $2\bar z+1=-iz-1$.
Now solve the system of equations
\begin{align} 2z-i\bar z&=-2\\ iz+2\bar z&=-2 \end{align}
to get $z$.
As an alternative to the fine method suggested, here another different way by $z=\rho e^{i\theta}$
$$2z-i\bar z=-2 \implies 2\rho e^{i\theta}-\rho ie^{-i\theta}=-2 \implies \rho =\frac{2}{ie^{-i\theta}-2e^{i\theta}}$$
which requires that denominator is real, that is
$$\Im \left(ie^{-i\theta}-2e^{i\theta}\right)=\cos \theta-2\sin \theta=0\implies \tan \theta =\frac12$$
and leads to $ie^{-i\theta}-2e^{i\theta}=\sin \theta-2\cos \theta=\frac3{\sqrt 5}$ such that
$$\rho =\frac{2}{ie^{-i\theta}-2e^{i\theta}} =\frac{2\sqrt 5}3$$
$$2z-i\bar z=-2$$ $$\left[\pmatrix{2&0\\0&2}-\pmatrix{0&-1\\1&0}\pmatrix{1&0\\0&-1}\right]X=\pmatrix{-2\\0}$$ $$\pmatrix{2&-1\\-1&2}X=\pmatrix{-2\\0}$$ $$X=\pmatrix{2&-1\\-1&2}^{-1}\pmatrix{-2\\0}$$ $$X=\pmatrix{-\frac43\\-\frac23}$$
Explanation of this method: We have an isomorphism of real vector spaces $\Bbb C\cong\Bbb R^2.$ Under this isomorphism, the conjugation map $\bar z:\Bbb C\to\Bbb C$ corresponds to the reflection linear transformation $\pmatrix{1&0\\0&-1}:\Bbb R^2\to\Bbb R^2$, multiplication by $i$ corresponds to the rotation linear transformation $\pmatrix{0&-1\\1&0}$, multiplication by $2$ correspons to $2I$.
Warning: $\bar z i$ and $i\bar z$ seem to give different matrices although they are equal in the field $\Bbb C$. We take the sequence of operations in consideration, I think.
