====== Question 3, Exercise 1.3 ======
Solutions of Question 3 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

====Question 3(i)====
Solve the quadratic equation: $\dfrac{1}{3} z^{2}+2 z-16=0$.

**Solution.**
Given
\begin{align}&\dfrac{1}{3}z^{2}+2 z-16=0\\
\implies &z^{2} + 6z - 48 = 0 \end{align}
Apply the quadratic formula:
$$ z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a},$$
where $$a = 1,\quad  b = 6,\quad \text{and}\quad  c = -48.$$
Then
\begin{align} 
z& = \dfrac{{-6 \pm \sqrt{36-4(1)(-48)}}}{2 \cdot 1} \\
& = \dfrac{{-6 \pm \sqrt{228}}}{2 \cdot 1} \\
&= \dfrac{{-6 \pm 2\sqrt{57}}}{2} \\
&= -3 \pm \sqrt{57} \end{align}
Hence Solution set $=\{ -3 \pm \sqrt{57} \}$.


====Question 3(ii)====
Solve the quadratic equation: $z^{2}-\frac{1}{2} z+17=0$.

**Solution.**

Given
$$ z^{2} - \frac{1}{2}z + 17 = 0 $$
Using the quadratic formula:
$$ z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} $$
Where\\
$$ a = 1, \quad b = -\dfrac{1}{2}, \quad c = 17 $$
Then
\begin{align}
z &= \dfrac{{-\left(-\dfrac{1}{2}\right) \pm \sqrt{{\left(-\dfrac{1}{2}\right)^2 - 4 \cdot 1 \cdot 17}}}}{2 \cdot 1} \\
&= \dfrac{{\dfrac{1}{2} \pm \sqrt{{\dfrac{1}{4} - 68}}}}{2} \\
&= \dfrac{{\dfrac{1}{2} \pm \sqrt{{\dfrac{1 - 272}{4}}}}}{2} \\
&= \dfrac{{\dfrac{1}{2} \pm \sqrt{{\dfrac{-271}{4}}}}}{2} \\
&= \dfrac{{\dfrac{1}{2} \pm \dfrac{\sqrt{-271}}{2}}}{2} \\
&= \dfrac{1 \pm \sqrt{271}i}{4}
\end{align}

Therefore, the solution set is:
$\left\{\dfrac{1 \pm \sqrt{271}i}{4}\right\} $

====Question 3(iii)====
Solve the quadratic equation: $z^{2}-6 z+25=0$.

**Solution.**

Given $$ z^{2} - 6z + 25 = 0 $$
Using the quadratic formula:
$$ z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} $$
Where $$ a = 1, \quad b = -6, \quad c = 25 $$
Then
\begin{align}
z &= \dfrac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 1 \cdot 25}}}}{2 \cdot 1} \\
&= \dfrac{{6 \pm \sqrt{{36 - 100}}}}{2} \\
&= \dfrac{{6 \pm \sqrt{{-64}}}}{2} \\
&= \dfrac{{6 \pm 8i}}{2} \\
&= 3 \pm 4i
\end{align}

Therefore, the solution set is:
$\{3 \pm 4i\}$

====Question 3(iv)====
Solve the quadratic equation: $z^{2}-9 z+11=0$.

**Solution.**

Given $$z^{2} - 9z + 11 = 0 $$
Using the quadratic formula:
$$ z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} $$

Where $$ a = 1, \quad b = -9, \quad c = 11 $$
Then
\begin{align}
z &= \dfrac{{-(-9) \pm \sqrt{{(-9)^2 - 4 \cdot 1 \cdot 11}}}}{2 \cdot 1} \\
&= \dfrac{{9 \pm \sqrt{{81 - 44}}}}{2} \\
&= \dfrac{{9 \pm \sqrt{37}}}{2}
\end{align}

Therefore, the solution set
$=\left\{ \dfrac{9 \pm \sqrt{37}}{2}\right\}$


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