Q2 TUT 5101

[Victor Ivrii]

Find the limit of each function at the given point $z_0$, or explain
why it does not exist.
\begin{equation*}
f(z)=\frac{z^3-8i}{z+2i},\quad z\ne 2i, \qquad\text{at   } z_0=2i.
\end{equation*}

[Yatong Yu]

f(z) = z³+(2i)³i/z+2i
      = (z+2i)(z²-2iz+(2i)²)/z+2i
      =z²-2iz-4
lim_z->2i f(z)=lim_z->2i z²-2iz-4
                  =(2i)²-2i(2i)-4
                  = -4+4-4
                  =-4

Practically  unreadable despite all insane html "mathematics".
$$\begin{aligned}f(z) &= \frac{z^3+(2i)^3i}{z+2i}\\
      &= \frac{(z+2i)(z^2-2iz+(2i)^2)}{z+2i}\\
      &=z^2-2iz-4\\
\lim_{z\to2i} f(z)&=\lim_{z\to2i}  z^2-2iz-4\\
                  &=(2i)^2-2i(2i)-4\\
                  &= -4+4-4\\
                  &=-4
\end{aligned}$$