Question 8, Exercise 1.2

Solutions of Question 8 of Exercise 1.2 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.

Write $|2 z-i|=4$ in terms of $x$ and $y$ by taking $z=x+i y$.

Solution.

Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & |2x+i(2y-1)|=4 \\ \implies & \sqrt{(2x)^2+(2y-1)^2}=4 \end{align} Squaring on both sides \begin{align} & (2x)^2+(2y-1)^2 = 16\\ \implies & 4x^2+4y^2-4y+1-16=0 \\ \implies & 4x^2+4y^2-4y-15=0, \end{align} as required.

Write $|z-1|=|\bar{z}+i|$ in terms of $x$ and $y$ by taking $z=x+i y$.

Solution. Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \begin{align} & |(x+iy)-1| = |(x-iy)+i| \\ \implies & |x-1+iy| = |x-i(y-1)| \\ \implies & \sqrt{(x-1)^2+y^2} = \sqrt{x^2+(y-1)^2} \end{align} Squaring both sides, we get \begin{align} & (x-1)^2 + y^2 = x^2 + (y-1)^2 \\ \implies & x^2 - 2x + 1 + y^2 = x^2 + y^2 - 2y + 1 \\ \implies & -2x = -2y \\ \implies & x = y \\ \implies & x - y =0, \end{align} as required.

Write $|z-4 i|+|z+4 i|=10$ in terms of $x$ and $y$ by taking $z=x+i y$.

Solution.

Given: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy) + 4i| = 10 \\ \implies & |x + i(y - 4)| + |x + i(y + 4)| = 10 \\ \implies & \sqrt{x^2 + (y - 4)^2} + \sqrt{x^2 + (y + 4)^2} = 10 \\ \implies & \sqrt{x^2 + (y - 4)^2} = 10 - \sqrt{x^2 + (y + 4)^2}. \end{align} Square both sides: \begin{align} & x^2 + (y - 4)^2 = 100 - 20\sqrt{x^2 + (y + 4)^2} + x^2 + (y + 4)^2 \\ \implies & y^2 - 8y + 16 = 100 - 20\sqrt{x^2 + (y + 4)^2} + y^2 + 8y + 16 \\ \implies & -8y = 100 - 20\sqrt{x^2 + (y + 4)^2} + 8y \\ \implies & 20\sqrt{x^2 + (y + 4)^2} = 100 +8y +8y \\ \implies & 20\sqrt{x^2 + (y + 4)^2} = 100 +16y \\ \implies & 5\sqrt{x^2 + (y + 4)^2} = 25 + 4y. \end{align} Square both sides: \begin{align} & \left(5\sqrt{x^2 + (y + 4)^2}\right)^2 = \left(25 + 4y\right)^2 \\ \implies & 25 \left(x^2 + (y + 4)^2\right) = 625 + 200y + 16y^2. \\ \implies & 25 \left(x^2 + y^2 + 8y + 16\right) = 625 + 200y + 16y^2 \\ \implies & 25x^2 + 25y^2 + 200y + 400 - 625 - 200y - 16y^2 = 0 \\ \implies & 25x^2 + 9y^2 - 225 = 0 \\ \implies & 25x^2 + 9y^2 = 225, \end{align} as required.

Write $\frac{1}{2} \operatorname{Re}(i \bar{z})=4$ in terms of $x$ and $y$ by taking $z=x+i y$.

Solution. Given: $$\dfrac{1}{2} Re(i \bar{z}) = 4.$$ Put $z = x + i y$, then $\bar{z} = x - i y$.

We have \begin{align} & \dfrac{1}{2}Re(i(x-iy)) = 4 \\ \implies & \dfrac{1}{2}Re(ix+y)) = 4 \\ \implies & \dfrac{1}{2} y=4 \\ \implies & y=8, \end{align} as required.

Write $lm\left(\dfrac{z-1}{2 i}\right)=-5$ in terms of $x$ and $y$ by taking $z=x+i y$.

Solution.

Given $$lm\left(\dfrac{z-1}{2 i}\right)=-5$$ Put $z=x+i y$, we have \begin{align} &lm\left(\dfrac{x+iy-1}{2 i}\right)=-5\\ \implies & lm\left(\dfrac{x-1+iy}{2 i}\times \dfrac{i}{i}\right)=-5 \\ \implies & lm\left(-\dfrac{i}{2}( x-1+iy)\right)=-5 \\ \implies & lm\left(-\dfrac{(x-1)}{2}i+\dfrac{y}{2}\right)=-5 \\ \implies & -\dfrac{x-1}{2}=-5 \\ \implies & x-1=10 \\ \implies & x=11, \end{align} as required.

Write $-2 \leq Im(z+i) \leq 3$ in terms of $x$ and $y$ by taking $z=x+i y$.

Solution. Given $$-2 \leq Im(z+i) \leq 3.$$ Put $z=x+i y$, we have \begin{align} &-2 \leq Im(x+iy+i) \leq 3\\ \implies & -2 \leq Im(x+i(y+1)) \leq 3 \\ \implies & -2 \leq y+1 \leq 3 \end{align} Adding $-1$ in the above inequalities $$-3\leq y \leq 2,$$ as required.

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