# 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.

## Question 5(i)

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

### Solution

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 $-\dfrac{1}{2}\pm\dfrac{\sqrt{11}}{2}i$.

## Question 5(ii)

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

### Solution

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 $\dfrac{1\pm\sqrt{5}}{2}$.

## Question 5(iii)

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

### Solution

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 $1\pm\sqrt{1-i}$.

## Question 5(iv)

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

### Solution

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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