Tomi's Challenges - #7

First, let's plug in $Re(z)$ and $Im(z)$ .

$\Large \lim_{y\to\infty} \left (\frac{Re(2021)+Im(y)+Im(2021\times z)}{Im(\frac{z^2+iz}{2})-\frac{2021}{2}} \right)^{y}$

Next, you need to know that $Re(z)$ and $Im(z)$ are both linear operators . This means they satisfy the following two properties:

• Additivity

$Re(a+b)=Re(a)+Re(b)$ $\newline$ $Im(a+b)=Im(a)+Im(b)$

• Homogeneity (Here, $\alpha$ is a real constant.)

$Re(\alpha a)=\alpha Re(a)$ $\newline$ $Im(\alpha a)=\alpha Im(a)$

It can be easily shown that they also satisfy the following properties: $\newline$

$Re(\alpha)=\alpha$ $\newline$ $Im(\alpha)=0$ $\newline$ $Re(iz)=-Im(z)$ $\newline$ $Im(iz)=Re(z)$ $\newline$

Using this, let's solve this exercise. We have:

$\Large \lim_{y\to\infty} \left (\frac{Re(2021)+Im(y)+Im(2021\times z)}{Im(\frac{z^2+iz}{2})-\frac{2021}{2}} \right)^{y}=$

$\Large \lim_{y\to\infty} \left (\frac{2021+0+2021Im(z)}{\frac{1}{2}Im(z^2+iz)-\frac{2021}{2}} \right)^{y}=$

$\Large \lim_{y\to\infty} \left (\frac{2021+2021y}{\frac{1}{2}Im(z^2)+\frac{1}{2}Im(iz)-\frac{2021}{2}} \right)^{y}=$

$\Large \lim_{y\to\infty} \left (\frac{2021+2021y}{\frac{1}{2}4042y+\frac{1}{2}Re(z)-\frac{2021}{2}} \right)^{y}=$

$\Large \lim_{y\to\infty} \left (\frac{2021+2021y}{2021y+\frac{2021}{2}-\frac{2021}{2}} \right)^{y}=$

$\Large \lim_{y\to\infty} \left (\frac{2021+2021y}{2021y} \right)^{y}=$

$\Large \lim_{y\to\infty} \left (\frac{1+y}{y} \right)^{y}=$

$\Large \lim_{y\to\infty} \left (1+\frac{1}{y} \right)^{y}=\boxed{e}$

So the answer is $\left \lfloor{40e}\right \rfloor=108$ .