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 ${{z}^{2}}+z+3=0$

${{z}^{2}}+z+3=0$
According to the quadratic formula, we have
$a=1,\,\,\,b=1$ and $c=3$
Quadratic formula is
$$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( 1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{-1\pm \sqrt{1-12}}{2}\\ z&=\dfrac{-1\pm \sqrt{11}}{2}i\\ z&=-\dfrac{1}{2}+\dfrac{\sqrt{11}}{2}i,-\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}i\end{align}$$

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

${{z}^{2}}-1=z$
${{z}^{2}}-z-1=0$
According to the quadratic formula, we have
$a=1,\quad b=-1$ and $c=-1$
Quadratic formula is
$$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{1\pm \sqrt{1+4}}{2}\\ z&=\dfrac{1\pm \sqrt{5}}{2}\\ z&=\dfrac{1+\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}\end{align}$$

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

${{z}^{2}}-2z+i=0$
According to the quadratic formula, we have
$a=1,\,\,\,b=-2$ and $c=i$
Quadratic formula is
$$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( i \right)}}{2\left( 1 \right)}\\ z&=\dfrac{2\pm \sqrt{4-4i}}{2}\\ z&=\dfrac{2\pm 2\sqrt{1-i}}{2}\\ z&=1\pm \sqrt{1-i}\\ z&=1+\sqrt{1-i},1-\sqrt{1-i}\end{align}$$

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

${{z}^{2}}+4=0$
According to the quadratic formula, we have
$a=1,\,\,\,b=0$ and $c=4$
Quadratic formula is
$$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( 0 \right)\pm \sqrt{{{\left( 0 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{\pm \sqrt{-16}}{2}\\ z&=\dfrac{\pm 4i}{2}\\ z&=\pm 2i\\ z&=2i,-2i\end{align}$$

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