====== 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 $\pm 2i$.

====Go To====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit01:ex1-3-p3 |< Question 3 & 4]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit01:ex1-3-p5|Question 6 >]]</btn></text>