====== Question 4, Exercise 1.1 ======
Solutions of Question 4 of Exercise 1.1 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 4(i)====

Find the values of real number $x$ and $y$ in each of the following: $(2+3i)x+(1+3i)y+2=0$

**Solution.** 

\begin{align}&(2+3i)x+(1+3i)y+2=0\\
\implies &(2x+y+2)+(3x+3y)i=0.\end{align}
Comparing real and imaginary parts
\begin{align}
2x+y+2&=0 \quad \cdots(1)\\
3x+3y&=0\quad \cdots (2)
\end{align}
From (2), 
\begin{align}
&3x=-3y \\
x=-y \quad ... (3) \end{align}
Putting value of $x$ in (1)
\begin{align}2(-y)+y+2&=0\\
-2y+y&=-2\\
-y&=-2\\
y&=2\end{align}
Putting in $(3)$, we have $x=-2$.

Hence $x=-2$ and $y=2$.
GOOD


 ====Question 4(ii)====

Find the values of real number $x$ and $y$ in each of the following: $\dfrac{x}{(1+i)}+\dfrac{y}{1-2i}=1$

**Solution.** 

\begin{align}&\dfrac{x}{(1+i)}+\dfrac{y}{1-2i}=1\\
\implies &\dfrac{x(1-2i)+y(1+i)}{(1+i)(1-2i)}=1\\
\implies &\dfrac{x-i2x+y+iy}{1-2i^2+i-2i}=1\\
\implies &\dfrac{(x+y)+(y-2x)i}{3-i}=1\\
\implies & (x+y)+(y-2x)i=3-i\\
\implies & (x+y-3)+(y-2x+1)i=0.\end{align}
Comparing real and imaginary parts
\begin{align}&x+y-3=0\cdots\cdots(1)\\
&y-2x+1=0\cdots\cdots(2) \end{align}
From $(2)$, we have
&y=2x-1\cdots \cdots (3)\end{align}
Put the value of $y$ in $(1)$
\begin{align}& x+(2x-1)-3=0\\
\implies &3x-4=0\\
\implies &x=\dfrac{4}{3}\end{align}
Put value of $x$ in $(3)$
\begin{align}&y=2(\dfrac{4}{3})-1\\
\implies &y=\dfrac{8}{3}-1\\
\implies &y=\dfrac{5}{3}\end{align}
Hence $x=\dfrac{4}{3}$ and $y=\dfrac{5}{3}$.
GOOD

====Question 4(iii)====

Find the values of real number $x$ and $y$ in each of the following: $\dfrac{x}{(2+i)}=\dfrac{1-5i}{(3-2i)}+\dfrac{y}{2-i}$

**Solution.** 

\begin{align}&\dfrac{x}{(2+i)}=\dfrac{(1-5i)(2-i)+y(3-2i)}{(3-2i)(2-i)}\\
\Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2+5i^2-10i-i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\
\Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2-5-11i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\
\Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(3y-3)+(-11-2y)i}{(4-7i)}\\
\Rightarrow \quad & x(4-7i)=((3y-3)+(-11-2y)i)(2+i)\\
\Rightarrow \quad & 4x-7xi=(3y-3)(2+i)+(-11i-2yi)(2+i)\\
\Rightarrow \quad & 4x-7xi=6y-6+3yi-3i-22i-4yi-11i^2-2yi^2\\
\Rightarrow \quad & 4x-7xi=6y-6+3yi-3i-22i-4yi+11+2y\\
\Rightarrow \quad & 4x-7xi=8y+5-25i-yi\\
\Rightarrow \quad & 4x-7xi=8y+5+(-25-y)i\end{align}
Now do yourself.
====Question 4(iv)====

Find the values of real number $x$ and $y$ in each of the following: $x(1+i)^2+y(2-i)^2=3+10i$

**Solution.** 

\begin{align}&x(1+i)^2+y(2-i)^2=3+10i\\
\Rightarrow \quad &x(1+i^2+2i)+y(4+i^2-4i)=3+10i\\
\Rightarrow \quad &x(1-1+2i)+y(4-1-4i)=3+10i\\
\Rightarrow \quad &2xi+y(3-4i)=3+10i\\
\Rightarrow \quad &3y+(2x-4y)i=3+10i\\
\Rightarrow \quad &3y+(2x-4y)i=3+10i\end{align}
Now do yourself

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