Question 4 & 5, Review Exercise 1

Solutions of Question 4 & 5 of Review Exercise 1 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.

If ${{z}_{1}}=2-i$, ${{z}_{2}}=1+i,$ find $\left|\dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}\right|$.

Given $z_1=2-i$ and $z_2=1+i$, so we have \begin{align} \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}&=\dfrac{\left( 2-i \right)+\left( 1+i \right)+1}{\left( 2-i \right)-\left( 1+i \right)+1}\\ &=\dfrac{2-i+1+i+1}{2-i-1-i+1}\\ &=\dfrac{4}{2-2i}\\ &=\dfrac{2}{1-i}\\ &=\dfrac{2}{1-i}\times \dfrac{1+i}{1+i}\\ &=\dfrac{2\left( 1+i \right)}{1+1}\\ &=\dfrac{2\left( 1+i \right)}{2}\\ &=1+i.\end{align} Now \begin{align}\left|\dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}\right|&=\sqrt{{{1}^{2}}+{{1}^{2}}}\\ &=\sqrt{2}\end{align}

Find the modulus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$.

\begin{align}\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}&=\dfrac{{{\left( 1+i \right)}^{2}}-{{\left( 1-i \right)}^{2}}}{1+1}\\ &=\dfrac{\left( 1+2i-1 \right)-\left( 1-2i-1 \right)}{2}\\ &=\dfrac{2i+2i}{2}\\ &=2i\end{align} Now \begin{align}\left|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\right|&=\sqrt{{{2}^{2}}}\\ &=\sqrt{4} = 2. \end{align}

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