Complex Numbers
Complex Number
nta_pyq_2025_jan
Grade 11
Question:
Let $z_{1},z_{2}$ and $z_{3}$ be three complex numbers on the circle $|z|=1$ with $\arg(z_{1})=-\dfrac{\pi}{4}$, $\arg(z_{2})=0$ and $\arg(z_{3})=\dfrac{\pi}{4}$. If $\bigl|z_{1}\overline{z_{2}}+z_{2}\overline{z_{3}}+z_{3}\overline{z_{1}}\bigr|^{2}=\alpha+\beta\sqrt{2}$, $\alpha,\beta\in\mathbb{Z}$, then the value of $\alpha^{2}+\beta^{2}$ is:
Step-by-Step Solution
Key Concept: On the unit circle, $z=e^{i\theta}$ and $\bar z = e^{-i\theta}$. Each $z_{j}\bar z_{k}=e^{i(\theta_{j}-\theta_{k})}$ collapses to a single exponential, making the sum easy to evaluate in Cartesian form.
On $|z|=1$, $z_{1}=e^{-i\pi/4},\ z_{2}=1,\ z_{3}=e^{i\pi/4}$.
$$z_{1}\bar z_{2}=e^{-i\pi/4}=\frac{1-i}{\sqrt{2}},$$
$$z_{2}\bar z_{3}=e^{-i\pi/4}=\frac{1-i}{\sqrt{2}},$$
$$z_{3}\bar z_{1}=e^{i\pi/2}=i.$$
Sum:
$$S = \frac{2(1-i)}{\sqrt{2}}+i = \sqrt{2}(1-i)+i = \sqrt{2}+i(1-\sqrt{2}).$$
$$|S|^{2} = 2+(1-\sqrt{2})^{2} = 2+1-2\sqrt{2}+2 = 5-2\sqrt{2}.$$
So $\alpha=5,\ \beta=-2$, hence $\alpha^{2}+\beta^{2}=25+4=29.$
Correct Answer: 2