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

Given: $$z^2-2z+5=0 \ldots (i)$$ Put $z=1+2i$ in equaiton (i), we have \begin{align}L.H.S.&=(1+2i)^2-2(1+2i)+5\\ &=1+(2i)^2+4i-2-4i+5\\ &=1-4+5\\ &=0\end{align} This implies $1+2i$ is solution of the given equation.

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