Question

The complex number $z=x+iy$ which satisfy the equation $\dfrac{|z-5i|}{|z+5i|}=1$, lie on$\begin{aligned} & (A)\text{ The x-axis}\text{.} \\\ & (B)\text{ The straight line y=5}\text{.} \\\ & (C)\text{ A circle passes through the origin}\text{.} \\\ & (D)\text{ None of these}\text{.} \\\ \end{aligned}$

Answer 1

Hint: We should place $z=x+iy$ in the given equation. We have to apply the concept of finding the magnitude of complex numbers to solve the problem. Now we have to find the equation of locus of $\left( x,y \right)$. Once the equation is obtained, we also have to obtain the properties of the equation of locus.

Complete step-by-step answer:
We know that for a complex number $z=x+iy$, the magnitude is equal to $|z|=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Given a complex number $z=x+iy$ . It is also given that $z=x+iy$ satisfy the equation $\dfrac{|z-5i|}{|z+5i|}=1$As $z=x+iy$ satisfies $\dfrac{|z-5i|}{|z+5i|}=1$, $\dfrac{|z-5i|}{|z+5i|}=1$ passes through $z=x+iy$.
Now let us consider $\dfrac{|z-5i|}{|z+5i|}=1$
By using cross multiplication,
$|z-5i|=|z+5i|......(1)$.
Now let us substitute $z=x+iy$in (1).
$|x+iy-5i|=|x+iy+5i|.....(2)$
Now let us Consider $|x+iy-5i|$
Now we will separate the real part of the complex number at one side and the imaginary part of the complex number at the other side.
Hence, $|x+iy-5i|=|x+i(y-5)|.........(3)$
Now let us consider $|x+iy+5i|$
Now we will separate the real part of the complex number at one side and the imaginary part of the complex number at the other side.
Hence, $|x+iy+5i|=|x+i(y+5)|.........(4)$
Now we will substitute equation (3) and equation (4) in equation (2)
$|x+i(y-5)|=|x+i(y+5)|......(5)$
Let us assume z1 as $x+i(y-5)$.
Hence, we get z1 as ${{z}_{1}}=x+i(y-5)...........(6)$
Let us assume z2 as $x+i(y+5)$
Hence, we get z2 as ${{z}_{2}}=x+i(y+5)...........(7)$.
We know that the magnitude of $z=a+ib$ is $|z|=|a+ib|=\sqrt{{{a}^{2}}+{{b}^{2}}}$.
By using the formula of magnitude of complex number $z=a+ib$,
We get Magnitude of z1 =$|{{z}_{1}}|$=$|x+i(y-5)|$
Here, the value of a is equal to x and the value of b is equal to (y-5).
${{z}_{1}}=|x+i(y-5)|=\sqrt{{{x}^{2}}+{{\left( y-5 \right)}^{2}}}.....(8)$
In the similar manner, by using the formula of magnitude of complex number $z=a+ib$,
We get magnitude of z2 = $|{{z}_{2}}|=|x+i(y+5)|$
Here, the value of x is equal to x and the value of b is equal to (y+5).
By applying the formula for magnitude of a complex number,
${{z}_{2}}=|x+i(y+5)|=\sqrt{{{x}^{2}}+{{\left( y+5 \right)}^{2}}}.....(9)$
Now we will Substitute equation (8) and equation (9) in equation (5)
$\sqrt{{{x}^{2}}+{{\left( y-5 \right)}^{2}}}=\sqrt{{{x}^{2}}+{{(y+5)}^{2}}}.....(10)$
In equation (5), we will square on both sides.
${{x}^{2}}+{{\left( y-5 \right)}^{2}}={{x}^{2}}+{{(y+5)}^{2}}$
$\Rightarrow {{x}^{2}}+{{y}^{2}}-10y+25={{x}^{2}}+{{y}^{2}}+10y+25$
$\Rightarrow \left( {{x}^{2}}+{{y}^{2}}-10y+25 \right)-\left( {{x}^{2}}+{{y}^{2}}+10y+25 \right)=0$
$\Rightarrow {{x}^{2}}+{{y}^{2}}-10y+25-{{x}^{2}}-{{y}^{2}}-10y-25=0$
$\Rightarrow -20y=0$
Now we will multiply with (-1) on both sides
$\Rightarrow 20y=0$
$\Rightarrow y=0$
We know that y=0 represents the equation of x-axis.
Hence, option (A) is correct.

Note: This problem can also get solved in another method.
Now we will multiply and divide $\dfrac{|z-5i|}{|z+5i|}$ with the conjugate of $|z-5i|$ i.e $|z+5i|$.
$\dfrac{|z-5i|}{|z+5i|}.\dfrac{|z+5i|}{|z+5i|}=1$
We know that $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ and ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
$\Rightarrow \dfrac{|{{z}^{2}}-{{(5i)}^{2}}|}{|z+5i{{|}^{2}}}=1$
By using cross multiplication
$\Rightarrow |{{z}^{2}}-{{(5i)}^{2}}|=|z+5i{{|}^{2}}......(1)$
Now we will substitute $z=x+iy$ in (1)

& \Rightarrow |{{\left( x+iy \right)}^{2}}-25{{i}^{2}}|={{\left( \sqrt{{{x}^{2}}+{{\left( y+5 \right)}^{2}}} \right)}^{2}}(Here:{{i}^{2}}=-1) \\\ & \Rightarrow |{{x}^{2}}-{{y}^{2}}+25+i(2xy)|={{x}^{2}}+{{y}^{2}}+10y+25 \\\ & \Rightarrow {{\left( {{x}^{2}}-{{y}^{2}}+25 \right)}^{2}}+4{{x}^{2}}{{y}^{2}}={{\left( {{x}^{2}}+{{y}^{2}}+10y+25 \right)}^{2}} \\\ & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}}+10y+25 \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+25 \right)}^{2}}=4{{x}^{2}}{{y}^{2}} \\\ & \Rightarrow \left( 2{{x}^{2}}+10y+25 \right)\left( 2{{y}^{2}}+10y \right)=4{{x}^{2}}{{y}^{2}} \\\ & \Rightarrow 4{{x}^{2}}{{y}^{2}}+20{{y}^{3}}+50{{y}^{2}}+20{{x}^{2}}y+100{{y}^{2}}+250y=4{{x}^{2}}{{y}^{2}} \\\ & \Rightarrow 20{{y}^{3}}+50{{y}^{2}}+20{{x}^{2}}y+100{{y}^{2}}+250y=0 \\\ & \Rightarrow y=0(or)2{{x}^{2}}+2{{y}^{2}}+15y+250=0 \\\ & \\\ \end{aligned}$$ The locus is x-axis (or) $$2{{x}^{2}}+2{{y}^{2}}+15y+250=0$$.