Question

# The locus of the point $z=x+iy$satisfying$\left| \dfrac{z-2i}{z+2i} \right|=1$isa) $x+iy=0\text{ then x=0, y=0}$b) $y-axis$c) $y=2$d) $x=2$

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Hint: Here we will first put the value of z in$\left| \dfrac{z-2i}{z+2i} \right|=1$. Then multiply the term with the conjugate $x-iy+2i\Rightarrow x-i(y+2)$. Then solve to find the value of $y$.

Complete step-by- step solution:
Given that: $z=x+iy$and equation is $\left| \dfrac{z-2i}{z+2i} \right|=1$
Here, $\left| \dfrac{z-2i}{z+2i} \right|=1$
We put $z=x+iy$in the equation$\left| \dfrac{z-2i}{z+2i} \right|=1$
Now, $\left| \dfrac{x+iy-2i}{x+iy+2i} \right|=1$
We multiply with the conjugate [$x-iy+2i\Rightarrow x-i(y+2)$]
$\Rightarrow \left| \dfrac{x+iy-2i}{x+iy+2i} \right|\left| \dfrac{x-iy-2i}{x-iy+2i} \right|=1$
Using identity [$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$] and removing$\left| {} \right|$- modulus
$\Rightarrow \sqrt{\dfrac{{{x}^{2}}+{{\left( y-2 \right)}^{2}}}{{{x}^{2}}+{{\left( y+2 \right)}^{2}}}}=1$
We, apply cross multiplication
$\Rightarrow \sqrt{{{x}^{2}}+{{\left( y-2 \right)}^{2}}}=\sqrt{{{x}^{2}}+{{\left( y+2 \right)}^{2}}}$
Now squaring each term on both the sides of the equation,
$\Rightarrow {{x}^{2}}+{{\left( y-2 \right)}^{2}}={{x}^{2}}+{{\left( y+2 \right)}^{2}}$
Now expand the equation using identity [${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$]
$\Rightarrow {{x}^{2}}+{{y}^{2}}+4-4y={{x}^{2}}+{{y}^{2}}+4+4y$
Now we solve and find the value of $y$:
$\Rightarrow 8y=0$$\therefore y=0$
As $y$ is $0$, it will lie on the$x-axis$.
The locus of the point $z=x+iy$satisfying$\left| \dfrac{z-2i}{z+2i} \right|=1$is $x-axis$
Hence, option A is the correct answer.

Note: In these types of problems, where the value is assigned and we need to check a defined condition for an equation - First, we need to put the assigned value in equation then use proper formulas to break it in the simplest form and compare the known and unknown terms. Also, for if any point lies on $x-axis$then its corresponding value on the $y-axis$ will be $0$and vice-versa.
When x, y are real numbers and $x+iy=0,\text{ then x=0 and y=0}$ When a, b, c and d are the real numbers and$a+ib=c+id$ then, $a=c\,\text{ and b=d}$. When the sum of the two complex numbers is the real and the product of the two complex numbers is also real then the given complex numbers are conjugate to each other.