Prove It!

May 24, 2022September 28, 2022 / Michael Xue / Leave a comment

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!

May 19, 2022October 11, 2022 / Michael Xue / Leave a comment

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!

May 17, 2022September 28, 2022 / Michael Xue / Leave a comment

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 $|z_2-z_3|=|z_1-z_3| = \sqrt{3}.$