====== 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. 

=====Question 3(i)=====
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.

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