# Prove It!

Given complex number $u, v$ and their product $uv$:

Show that the triangle formed by $uv, v$ and $uv-v$ is similar to that formed by $u, 1$ and $u-1$.

Here is my proof:

From Fig. 1, we see that $\angle COB = \theta_{uv}, \angle DOB = \theta_v, \angle AOB = \theta_u.$

It follows that $\angle COD = \angle COB - \angle DOB = \theta_{uv}-\theta_v \overset{\theta_{uv} = \theta_u + \theta_v}{=} \theta_u + \theta_v - \theta_v = \theta_u = \angle AOB.$

And, $\frac{|u\cdot v|}{|u|} \overset{|u\cdot v|=|u|\cdot|v|}{=} \frac{|u|\cdot|v|}{|u|} = |v| =\frac{|v|}{1} = \frac{|v|}{|1|}\implies \frac{OC}{OA}=\frac{OD}{OB}.$

Therefore, $\left(\angle COD = \angle AOB, \frac{OC}{OA} = \frac{OD}{OB}\right) \implies \triangle OCD \sim \triangle OAB.$

# Integrate This!

Evaluate $\displaystyle\int \frac{(x-1)^2 e^x}{(x^2+1)^2}\; dx$ $\displaystyle\int \frac{(x-1)^2 e^x}{(x^2+1)^2}\; dx$ $= \displaystyle\int \frac{(x^2-2x+1)e^x}{(x^2+1)^2}\;dx$ $= \displaystyle\int \frac{(x^2+1) e^x-2x e^x}{(x^2+1)^2}\;dx$ $= \displaystyle\int \frac{(x^2+1)(e^x)'-(x^2+1)' e^x}{(x^2+1)^2}\;dx$ $= \displaystyle\int \left(\frac{e^x}{x^2+1}\right)'\;dx$ $= \frac{e^x}{x^2+1}$

# Arrows Arrows and more Arrows!

Given $\begin{cases} z_1+z_2+z_3 =0 \quad\quad\quad\quad\quad(1-1)\\ |z_1|=|z_2|=|z_3|=1\quad\quad\quad(1-2)\end{cases}$

where $z_1, z_2, z_3$ are complex numbers.

Prove that $z_1, z_2, z_3$ are the vertices of an equilateral triangle inscribed in a circle.

We prove $|z_1-z_2|=|z_2-z_3| = |z_1-z_3|$ as follows: $|z_1-z_2|^2=(z_1-z_2)(\overline{z_1-z_2})$ $= (z_1-z_2)(\overline{z_1}-\overline{z_2})$ $= z_1\overline{z_1}-z_1\overline{z_2}-z_2\overline{z_1}+z_2\overline{z_2}.$

By (1-2), $z_1\overline{z_1} = |z_1|^2=1 \implies z_1\overline{z_1} =1, z_2\overline{z_2} = |z_2|^2=1 \implies z_2\overline{z_2} =1.$ $= 1-z_1\overline{z_2}-z_2\overline{z_1}+1$ $= 2-(z_1\overline{z_2} + z_2\overline{z_1}).\quad\quad\quad(2-1)$

Since (1-1) gives $z_3 = -(z_1+z_2) \implies \overline{z_3}=-(\overline{z_1+z_2})=-(\overline{z_1} + \overline{z_2}),$ we have $z_3\overline{z_3} = -(z_1+z_2)(-(\overline{z_1}+\overline{z_2}))$ $= (z_1+z_2)(\overline{z_1}+\overline{z_2})$ $= z_1\overline{z_1} + z_1\overline{z_2}+z_2\overline{z_1} + z_2\overline{z_2}$ $= 1 + z_1\overline{z_2}+z_2\overline{z_1} + 1$ $= 2 + z_1\overline{z_2}+z_2\overline{z_1}.$

That is, $z_3\overline{z_3} = 2 + z_1\overline{z_2}+z_2\overline{z_1}.$

From $z_3\overline{z_3} = |z_3|^2=1$, we see that $1 = 2 + z_1\overline{z_2}+z_2\overline{z_1}\implies z_1\overline{z_2}+z_2\overline{z_1}=-1.\quad\quad\quad(2-2)$

Substituting (2-2) into (2-1) gives $|z_1-z_2| = 2-(-1) = \sqrt{3}.$

Similarly, it can be shown that $|z_2-z_3| = |z_1-z_3| = \sqrt{3}.$

Hence, $|z_1-z_2| = |z_2-z_3| = |z_1-z_3|.$

i.e.,

Triangle with vertices $z_1, z_2, z_3$ is equilateral.

Exercise-1 Show that $|z_2-z_3|=|z_1-z_3| = \sqrt{3}.$