Question 2 & 3, Review Exercise 1

Solutions of Question 2 & 3 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.

Show that ${{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}=0$, $\forall n\in N$

\begin{align}{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}&=0\\ L.H.S.&={{i}^{n}}+{{i}^{n}}\cdot i+{{i}^{n}}\cdot {{i}^{2}}+{{i}^{n}}\cdot {{i}^{3}}\\ &={{i}^{n}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)\\ &={{i}^{n}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)\\ &=i\left( 1+i+{{\left( i \right)}^{2}}+i{{\left( i \right)}^{2}} \right)\\ &=i\left( 1+i+\left( -1 \right)+i\left( -1 \right) \right)\\ &=i\left( 1+i-1-i \right)\\ &=i\left( 0 \right)\\ &=0=R.H.S.\end{align}

Express the complex number $\left( 1+3i \right)+\left( 5+7i \right)$ in the form of $x+iy$.

$\left( 1+3i \right)+\left( 5+7i \right)=1+5+3i+7i$ $=6+10i$

Express the complex number $\left( 1+3i \right)-\left( 5+7i \right)$ in the form of $x+iy$.

\begin{align}\left( 1+3i \right)-\left( 5+7i \right)&=1+3i-5-7i\\ &=1-5+3i-7i\\ &=-4-4i\end{align}

Express the complex number $\left( 1+3i \right)\left( 5+7i \right)$ in the form of $x+iy$.

\begin{align}(1+3i)(5+7i)&=5+7i+15i+21{{i}^{2}}\\ &=5-21+7i+15i\\ &=-16+22i\end{align}

Express the complex number $\dfrac{1+3i}{5+7i}$ in the form of $x+iy$.

\begin{align}\dfrac{1+3i}{5+7i}&=\dfrac{1+3i}{5+7i}\times \dfrac{5-7i}{5-7i}\\ &=\dfrac{5-7i+15i+21}{25+49}\\ &=\dfrac{26+8i}{74}\\ &=\dfrac{13+4i}{37}\\ &=\dfrac{13}{37}+\dfrac{4}{37}i\end{align}

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