# Question 3 & 4, Exercise 1.3

Solutions of Question 3 & 4 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 3

Show that each ${{z}_{1}}=-1+i$ and ${{z}_{2}}=-1-i$ satisfied the equation ${{z}^{2}}+2z+2=0$

### Solution

Given:
$$z^2+2z_1+2=0\quad \ldots (i)$$
Put the value of $z_1=-1+i$ in (i)
\begin{align}L.H.S &= (-1+i)^2+2(-1+i)+2\\
&=1-2i-1-2+2i+2\\
&=0=R.H.S\end{align}
This implies $z_1=-1+i$ satisfied the given equation.

Now put $z_2=-1-i$ in (i)
\begin{align}
L.H.S&=(-1-i)^2+2(-1-i)+2\\
&=1+2i-1-2-2i+2\\
&=0=R.H.S\end{align}
This implies $z_2=-1-i$ satisfied the equation.

## Question 4

Determine weather $1+2i$ is a solution of ${{z}^{2}}-2z+5=0$

### Solution

${{z}^{2}}-2z+5=0$

According to the quadratic formula, we have

$a=1,\quad b=-2$ and $c=5$

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( 5 \right)}}{2\left( 1 \right)}\\
z&=\dfrac{2\pm \sqrt{4-20}}{2}\\
z&=\dfrac{2\pm \sqrt{-16}}{2}\\
z&=\dfrac{2\pm 4i}{2}\\
z&=1\pm 2i\\
z&=1+2i,1-2i\end{align}

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