# Question 3, Exercise 1.4

Solutions of Question 3 of Exercise 1.4 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

If $\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \ldots\left(x_{n}+i y_{n}\right)=a+i b$, show that:

(i) $\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)\left(x_{3}^{2}+y_{3}^{2}\right) \ldots\left(x_{n}^{2}+y_{n}^{2}\right)=a^{2}+b^{2}$

(ii) $\sum_{r=1}^{n} \tan ^{-1}\left(\frac{y_{r}}{x_{r}}\right)=\tan ^{-1}\left(\frac{b}{a}\right)+2 k \pi, k \in \mathbb{Z}$

Solution.

Let $z_r=x_r+iy_r$, $r=1,2,...,n$ and $z=a+ib$. Then

\begin{align*} &|z_r|=\sqrt{x_r^2+y_r^2} \quad \text{and}\quad |z|=\sqrt{a^2+b^2}. \\ &\theta_k = \arg(z_k) = \tan^{-1}\left(\dfrac{y_r}{x_r}\right) \quad \text{and}\quad \theta=\tan^{-1}\left(\dfrac{b}{a}\right). \end{align*} We can write these complex numbers in polar form as: \begin{align*} z_r=|z_r| e^{i\theta_k} \quad \text{and}\quad z=|z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right)\ldots \left(x_{n}+i y_{n}\right)=a+i b\\ \implies &z_1 \cdot z_2 \cdot z_3 \cdots z_n = z. \end{align*} By using $(1)$, we have \begin{align*} &|z_1| e^{i\theta_1}\cdot |z_2| e^{i\theta_2} \cdot |z_3| e^{i\theta_3} \cdots |z_n| e^{i\theta_n}=|z| e^{i\theta} \\ \implies & \left(|z_1|\cdot|z_2|\cdot|z_3|\cdots |z_n|\right)e^{i(\theta_1+\theta_2+\theta_3+\ldots+\theta_n)} = |z| e^{i\theta}. \end{align*} This gives $$|z_1|\cdot|z_2|\cdot|z_3|\cdots |z_n| = |z| \,\, -- (2)$$ and $$\theta_1+\theta_2+\theta_3+\ldots+\theta_n = \theta + 2k\pi, \quad \text{where } k\in \mathbb{Z}.\,\, -- (3)$$ Taking square on both sides of $(2)$, we get $$|z_1|^2\cdot|z_2|^2\cdot|z_3|^2\cdots |z_n|^2 = |z|^2$$ implies $$\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)\left(x_{3}^{2}+y_{3}^{2}\right) \ldots\left(x_{n}^{2}+y_{n}^{2}\right)=a^{2}+b^{2}\,\,-- (4)$$ Now from $(3)$, we have $$\sum_{r=1}^{n} \theta_r = \theta + 2k\pi, \quad \text{where } k\in \mathbb{Z},$$ implies $$\sum_{r=1}^{n} \tan ^{-1}\left(\frac{y_{r}}{x_{r}}\right)=\tan ^{-1}\left(\frac{b}{a}\right)+2 k \pi, k \in \mathbb{Z}. \,\, -- (5)$$ $(4)$ and $(5)$ are our required results.

Alternative Method for Part (i)

We have given \begin{align*} & \left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \ldots\left(x_{n}+i y_{n}\right)=a+i b\\ \implies &\left| (x_{1} + i y_{1})(x_{2} + i y_{2})(x_{3} + i y_{3}) \ldots (x_{n} + i y_{n}) \right| = |a + i b|\\ \implies &|x_{1} + i y_{1}| \cdot |x_{2} + i y_{2}| \cdot |x_{3} + i y_{3}| \ldots |x_{n} + i y_{n}| = |a + i b|\\ \end{align*} Taking square on both side, we have \begin{align*} &|x_{1} + i y_{1}|^2 \cdot |x_{2} + i y_{2}|^2 \cdot |x_{3} + i y_{3}|^2 \ldots |x_{n} + i y_{n}|^2 = |a + i b|^2\\ \implies &\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)\left(x_{3}^{2}+y_{3}^{2}\right) \ldots\left(x_{n}^{2}+y_{n}^{2}\right)=a^{2}+b^{2}. \end{align*} As required.