2025-Q2e

L3CALC

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QUESTION

Given that: $\text{arg}\left(\frac{d+6i}{1-di}\right)=\frac{\pi }{4}$ find the possible value(s) of $d$, where $d$ is a real constant.

SOLVING

First, I decided to ratioanlize the complex number since I can see an imaginary number in the denominator.

TIP: If you see a root/imaginary number in the denominator, always try to rationalize first.

$\text{arg}\left(\frac{d+6i}{1-di}\times \frac{1+di}{1+di}\right)=\frac{\pi }{4}$

$\text{arg}\left(\frac{(d+6i)(1+d{i}_{})}{1+d^2}\right)=\frac{\pi }{4}$

$\text{arg}\left(\frac{d+d^{2}i+6i+6d{i}^2}{d^2+1}\right)=\frac{\pi }{4}$

$\text{arg}\left(\frac{d^{2}i+6i-5d}{d^2+1}\right)=\frac{\pi }{4}$

$\text{arg}\left(\frac{-5d+i(d^2+6)}{d^2+1}\right)=\frac{\pi }{4}$

Then, I decided to change it to a form of $a+bi$:

$\text{arg}\left(\frac{-5d}{d^2+1}+\frac{d^2+6}{d^2+1}i\right)=\frac{\pi }{4}$

I got stuck here. I have spent over 2 weeks trying to solve this question but kept failing.
A few weeks later from that stage, I made no progress until I was at church on Sunday, and suddenly came to my mind.

“If the argument is $\frac{\pi }{4}$, which is 45 degrees, wouldn’t $a$ and $b$ have to be equal the same???”

I was in the middle of a church service so I couldn’t do anything, however as soon as service finished, I rushed to grab a pen and paper. Fortunately, my thought was correct. Making $a=b$ would return a quadratic function which returns two negative numbers.

$\frac{-5d}{d^2+1}=\frac{d^2+6}{d^2+1}$

$-5d=d^2+6$

$d^2+5d+6=0$

$d=-2,or-3$

Therefore, doing all these would let me get a score of E8.