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.

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 $y=2$.

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 $y=\dfrac{5}{3}$.

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.

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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