Question

The complex number $ z=x+iy $ which satisfy the equation $\left| \dfrac{z-5i}{z+5i} \right|=1$, lies on:
  (a) x axis.
  (b) The straight line y=5.
  (c) A circle passing through the origin.
  (d) None of the above.

Answer 1

Hint: At first put the value of z in the given equation. Then try to remove the modulus by using the following formula:
$\left| a+ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$
Then simplify the equation to get the required answer.

Complete step-by-step solution:
In this problem we have a complex number z and $z=x+iy$. Here x is known as the real part of a complex number and y is known as the imaginary part of a complex number.
It is given in the question that z satisfies the following equation:
$\left| \dfrac{z-5i}{z+5i} \right|=1.......(1)$
Now let us put the value of z in equation (1).
$\Rightarrow \left| \dfrac{\left( x+iy \right)-5i}{\left( x+iy \right)+5i} \right|=1$
$\Rightarrow \left| \dfrac{x+i\left( y-5 \right)}{x+i\left( y+5 \right)} \right|=1$
We know that, $\left| \dfrac{a}{b} \right|=\dfrac{\left| a \right|}{\left| b \right|}$
Therefore we will get,
$\Rightarrow \dfrac{\left| x+i\left( y-5 \right) \right|}{\left| x+i\left( y+5 \right) \right|}=1$
Now we need to remove the modulus from the left hand side. To remove the modulus we will apply the following formula:
$\left| a+ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$
Therefore,
$\Rightarrow \dfrac{\sqrt{{{x}^{2}}+{{\left( y-5 \right)}^{2}}}}{\sqrt{{{x}^{2}}+{{\left( y+5 \right)}^{2}}}}=1$
Now we can do cross multiplication. Therefore,
$\Rightarrow \sqrt{{{x}^{2}}+{{\left( y-5 \right)}^{2}}}=\sqrt{{{x}^{2}}+{{\left( y+5 \right)}^{2}}}$
By squaring both sides of the equation we will get,
$\Rightarrow {{\left( \sqrt{{{x}^{2}}+{{\left( y-5 \right)}^{2}}} \right)}^{2}}={{\left( \sqrt{{{x}^{2}}+{{\left( y+5 \right)}^{2}}} \right)}^{2}}$
$\Rightarrow {{x}^{2}}+{{\left( y-5 \right)}^{2}}={{x}^{2}}+{{\left( y+5 \right)}^{2}}$
We can cancel out the similar terms from both sides of the equation.
$\Rightarrow {{\left( y-5 \right)}^{2}}={{\left( y+5 \right)}^{2}}$
$\Rightarrow {{y}^{2}}-10y+25={{y}^{2}}+10y+25$
By cancelling out the similar terms from both sides we will get,
$\Rightarrow -10y=10y$
$\Rightarrow 10y+10y=0$
$\Rightarrow 20y=0$
Divide the both sides of the equation by 20.
$\Rightarrow y=\dfrac{0}{20}$
$\Rightarrow y=0$
Therefore the complex number z satisfies the equation y=0. That means the imaginary part of the complex number is always zero.
Hence the complex number z lies on the x axis.
Therefore option (a) is correct.

Note: Before applying the modulus formula always separate the real part and the imaginary part of a complex number. Otherwise we will get a wrong answer.