Question

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

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