Question 5, Exercise 1.3

Solutions of Question 5 of Exercise 1.3 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.

Find the solutions of the equation ${{z}^{2}}+z+3=0$.

Given: $${{z}^{2}}+z+3=0.$$ According to the quadratic formula, we have
$a=1$, $b=1$ and $c=3$.
Thus we have $$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\ &=\dfrac{-1\pm \sqrt{1-12}}{2}\\ &=\dfrac{-1\pm \sqrt{11}}{2}i\end{align}$$ Thus the solutions of the given equation are $-\dfrac{1}{2}\pm\dfrac{\sqrt{11}}{2}i$.

Find the solutions of the equation ${{z}^{2}}-1=z$.

Given: $${{z}^{2}}-1=z$$ $$\implies {{z}^{2}}-z-1=0$$ According to the quadratic formula, we have
$a=1$, $b=-1$ and $c=-1$
So we have $$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)}\\ &=\dfrac{1\pm \sqrt{1+4}}{2}\\ &=\dfrac{1\pm \sqrt{5}}{2}.\end{align}$$ Thus the solutions of the given equations are $\dfrac{1\pm\sqrt{5}}{2}$.

Find the solutions of the equation ${{z}^{2}}-2z+i=0$

Given: $${{z}^{2}}-2z+i=0$$ According to the quadratic formula, we have
$a=1$, $b=-2$ and $c=i$
So we have $$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( i \right)}}{2\left( 1 \right)}\\ &=\dfrac{2\pm \sqrt{4-4i}}{2}\\ &=\dfrac{2\pm 2\sqrt{1-i}}{2}\\ &=1\pm \sqrt{1-i}\end{align}$$ Thus the solutions of the given equations are $1\pm\sqrt{1-i}$.

Find the solutions of the equation ${{z}^{2}}+4=0$

Given: $${{z}^{2}}+4=0$$ $$\begin{align} \implies &z^2-(2i)^2=0\\ \implies &(z+2i)(z-2i)=0\\ \implies &z+2i=0 \quad \text{or} \quad z-2i=0\\ \implies &z=-2i \quad \text{or} \quad z=2i.\end{align}$$ Thus, the solutions of the given equations are $\pm 2i$.

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