Finding a limit for a complex function

I have this complex limit which is rather difficult to solve.

\begin{equation} \lim_{z\longrightarrow3i}\frac{z^2+9}{z-3i} \end{equation}

clearly, it I insert $3i$ for $z$, I get $\frac{0}{0}$, so that won't work. I multiply with the complex conjugate, and get

\begin{equation} \frac{z^2+9}{z-3i}\frac{z+3i}{z+3i}=\frac{z(z^2+3i+9)+27i}{z^2+9} \end{equation}

But here I still get the form $\frac{0}{0}$ when I insert for $3i$. What is the right thing to do here?

Thanks

Solution 1:

As your limit has a look $\bbox[lightgreen]{\lim\frac{0}{0}}$, therefore, you can use the L'Hopital's rule:

\begin{equation} \lim_{z\longrightarrow{3i}}\frac{(z^2+9)'}{(z-3i)^{'}}=\lim_{z\longrightarrow{3i}}(2z)=6i. \end{equation}

Solution 2:

Hint. $z^2+9=z^2-9i^2=(z+3i)(z-3i)$.