What are the points of differentiability of $f(z)=\vert z \vert^2+\iota \bar z +1$?

Question

The function $f(z)=\vert z \vert^2+\iota \bar z +1$ is differentiable at

$1)\iota$

$2)1$

$3)-\iota$

$4)$No point in $\mathbb C$

Solution: $f(z)=\vert z \vert^2+\iota \bar z +1=z\bar z+\iota \bar z +1$

If $f$ is differentiable at $z_0\implies f$ satisfies Cauchy-Riemann equations at $z_0\implies \frac{\partial f}{\partial \bar z}=0 $ at $z=z_0\implies z+\iota=0\implies z=-\iota$.So, $f$ is differentiable at $z=-\iota$ only.

Now,$lim_{z\rightarrow -\iota}\frac{f(z)-f(-\iota)}{z+\iota}=lim_{z\rightarrow -\iota} \frac{z\bar z+\iota \bar z +1-1}{z+\iota}=lim_{z\rightarrow -\iota}\frac{\bar z(z+\iota)}{z+\iota}=lim_{z\rightarrow -\iota} \bar z=\iota$,.

But in answer key it is given that $f$ is differentiable at no point in $\mathbb C$. Where i'm doing wrong?

Answer 1

First of all, the standard notation is $i$, not $\iota$.

Other than that, you are right. The function is differentiable at $-i$ and only there. And $f'(-i)=i$.