BdMO National Higher Secondary 2020 P7

[Mursalin]

\(f\) হলো জটিল সংখ্যার সেটের ওপরে একটা ফাংশন যেন \(f(z)=\frac{1}{z^*}\) হয়, যেখানে \(z^*\) হলো \(z\)-এর জটিল অনুবন্ধী। \(S\) হলো ওইসব জটিল সংখার সেট যেন \(f(z)\)-এর বাস্তব অংশ \(\frac{1}{2020}\) আর \(\frac{1}{2020}\)-এর মধ্যে থাকে। যদি আমরা \(S\)-কে জটিল তলের একটা উপসেট হিসেবে বিবেচনা করি, তাহলে \(S\)-এর ক্ষেত্রফলকে \(m\pi\) আকারে লেখা যাবে যেখানে \(m\) একটা পূর্ণসংখ্যা। \(m\)-এর মান কত? (নোট: জটিল তল হলো কার্তেসীয় তলের মতোই যেখানে \(x\)-অক্ষকে বাস্তব অক্ষ আর \(y\)-অক্ষকে কাল্পনিক অক্ষ বলা হয়।)

মনে রেখো, একটা জটিল সংখ্যা হলো \(a+bi\) আকারের একটা সংখ্যা যেখানে \(a\) আর \(b\) বাস্তব সংখ্যা এবং \(i\) হলো এমন একটা সংখ্যা যেন \(i^2=-1\) হয়। যদি \(z=a+bi\) হয়, তাহলে \(z\)-এর জটিল অনুবন্ধী হলো \(z^*=a-bi\)।


\(f\) is a function on the set of complex numbers such that \(f(z)=\frac{1}{z^*}\), where \(z^*\) is the complex conjugate of \(z\). \(S\) is the set of complex numbers \(z\) such that the real part of \(f(z)\) lies between \(\frac{1}{2020}\) and \(\frac{1}{2018}\). If \(S\) is treated as a subset of the complex plane, the area of \(S\) can be expressed as \(m\pi\) where \(m\) is an integer. What is \(m\)? (Note: the complex plane is just like the Cartesian plane, where the \(x\)-axis is renamed as the real-axis and the \(y\)-axis is renamed as the imaginary axis.)

Remember, a complex number is a number of the form \(a+bi\), where \(a\) and \(b\) are real numbers and \(i\) is a number such that \(i^2=-1\). If \(z=a+bi\), the complex conjugate of \(z\) is \(z^*=a-bi\).

[Mehrab4226]

It actually took me a lot of time. Because I had to newly learn complex numbers(Because I didn't know how it works). But complex numbers had very little impact on the solution.

Let, $z=a+bi \in S$
Then, $f(z) = \frac{1}{a-bi}$
Or, $f(z) = \frac{(a+bi)}{(a+bi)(a-bi)} = \frac{a+bi}{a^2+b^2}$
Or, $f(z) = \frac{a}{a^2+b^2} + \frac{b}{a^2+b^2}i$
Thus the real part of $f(z)$ is $\frac{a}{a^2+b^2}$

Now we get two inequalities,
$\frac{a}{a^2+b^2} > \frac{1}{2020}$, or, $\frac{a^2+b^2}{a} < 2020$[$a$ cannot be negative] $\cdots (1)$

$\frac{a}{a^2+b^2} < \frac{1}{2018}$, Or, $\frac{a^2+b^2}{a} > 2018$ $\cdots(2)$

Comparing Both of them with the equation of the circle, from now on complex numbers are useless(in my solution at least) so x is the real part and y is the imaginary part from now.

$$x^2+y^2-2ax-2by+a^2+b^2-r^2 = 0$$

We will get a circular region (1) with area = $1010^2 \pi$
We will get a circular region (2) with area = $1009^2 \pi$

Now everything inside the circle (1) are solutions of inequality (1)
Everything outside the circle(2) are solutions of the inequality (2)
Since (2) is entirely inside (1) they have a common area $= 1010^2 \pi- 1009^2 \pi = 2019 \pi$
Thus m = 2019 (Ans)

I wanted to give the picture but the circles are too big and the common area is too small to differentiate by zoomed out picture.