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.

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

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.

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

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