Question 6, Exercise 1.2

Solutions of Question 6 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Show that for all complex numbers ${{z}_{1}}$and ${{z}_{2}}$. Verify that $|{{z}_{1}}{{z}_{2}}|=|{{z}_{1}}||{{z}_{2}}|$.

Suppose ${{z}_{1}}=a+bi$ and ${{z}_{2}}=c+di$. Then $|z_1=\sqrt{a^2+b^2}|$ and $|z_2=\sqrt{c^2+d^2}|$.
Now \begin{align} L.H.S.&=|{{z}_{1}}{{z}_{2}}|\\ &=|(a+bi)(c+di)|\\ &=|ac-bd+(ad+bc)i|\\ &=\sqrt{{{(ac-bd)}^{2}}+{{(ad+bc)}^{2}}}\\ &=\sqrt{{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}-2abcd+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+2abcd}\\ &=\sqrt{{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}}\\ &=\sqrt{{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}}\\ &=\sqrt{{{a}^{2}}({{c}^{2}}+{{d}^{2}})+{{b}^{2}}({{c}^{2}}+{{d}^{2}})}\\ &=\sqrt{({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})}\\ &=\sqrt{{{a}^{2}}+{{b}^{2}}}\sqrt{{{c}^{2}}+{{d}^{2}}}\\ &=|{{z}_{1}}||{{z}_{2}}|=R.H.S. \end{align}

Alternative Method
We know $|z|^2=z\bar{z}$, so we have \begin{align}|{{z}_{1}}{{z}_{2}}{{|}^{2}}&={{z}_{1}}{{z}_{2}}\overline{({{z}_{1}}{{z}_{2}})}\\ &={{z}_{1}}{{z}_{2}} \bar{z}_1 \bar{z}_2 \quad \because \overline{{{z}_{1}}{{z}_{2}}}=\bar{z}_1 \bar{z}_2\\ &=\left( {{z}_{1}}\overline{{{z}_{1}}} \right)\left( {{z}_{2}}\overline{{{z}_{2}}} \right)\\ &=|{{z}_{1}}{{|}^{2}}|{{z}_{2}}{{|}^{2}}\\ \Rightarrow \,\,\,\,|{{z}_{1}}{{z}_{2}}|&=|{{z}_{1}}||{{z}_{2}}|\end{align} proved.

Show that for all complex numbers ${{z}_{1}}$and ${{z}_{2}}$that $\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{|{{z}_{1}}|}{|{{z}_{2}}|}$, where ${{z}_{2}}\ne 0$

Suppose $z=a+bi$, then $|z|=\sqrt{a^2+b^2}$. We take

\begin{align}\left| \dfrac{1}{z} \right|&=\left| \dfrac{1}{a+bi} \right|\\ &=\left| \dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi} \right|\\ &=\left| \dfrac{a-bi}{{{a}^{2}}-{{(bi)}^{2}}} \right|\\ &=\left| \dfrac{a}{{{a}^{2}}+{{b}^{2}}}-\dfrac{b}{{{a}^{2}}+{{b}^{2}}}i \right|\\ &=\sqrt{{{\left( \dfrac{a}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}+{{\left( \dfrac{b}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}}\\ &=\sqrt{\dfrac{{{a}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}+\dfrac{{{b}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}}\\ &=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}}\\ &=\sqrt{\dfrac{1}{{{a}^{2}}+{{b}^{2}}}}\\ &=\dfrac{1}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\dfrac{1}{|z|}\\ \implies \left|\dfrac{1}{z} \right|&=\dfrac{1}{|z|} … (1)\end{align} Now \begin{align}L.H.S.&=\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\left| {{z}_{1}}\cdot \dfrac{1}{{{z}_{2}}} \right|\\ &=|{{z}_{1}}|\left| \dfrac{1}{{{z}_{2}}} \right|\\ &=|{{z}_{1}}|\cdot \dfrac{1}{|{{z}_{2}}|}, \text{ by using (1)}\\ &=\dfrac{|{{z}_{2}}|}{|{{z}_{2}}|}=R.H.S.\end{align}