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Question: The complex number \( z=x+iy \) which satisfy the equation \(\left| \dfrac{z-5i}{z+5i} \right|=1\), ...

The complex number z=x+iyz=x+iy which satisfy the equation ∣z−5iz+5i∣=1\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.

Explanation

Solution

Hint: At first put the value of z in the given equation. Then try to remove the modulus by using the following formula:
∣a+ib∣=a2+b2\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+iyz=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:
∣z−5iz+5i∣=1.......(1)\left| \dfrac{z-5i}{z+5i} \right|=1.......(1)
Now let us put the value of z in equation (1).
⇒∣(x+iy)−5i(x+iy)+5i∣=1\Rightarrow \left| \dfrac{\left( x+iy \right)-5i}{\left( x+iy \right)+5i} \right|=1
⇒∣x+i(y−5)x+i(y+5)∣=1\Rightarrow \left| \dfrac{x+i\left( y-5 \right)}{x+i\left( y+5 \right)} \right|=1
We know that, ∣ab∣=∣a∣∣b∣\left| \dfrac{a}{b} \right|=\dfrac{\left| a \right|}{\left| b \right|}
Therefore we will get,
⇒∣x+i(y−5)∣∣x+i(y+5)∣=1\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:
∣a+ib∣=a2+b2\left| a+ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}
Therefore,
⇒x2+(y−5)2x2+(y+5)2=1\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,
⇒x2+(y−5)2=x2+(y+5)2\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,
⇒(x2+(y−5)2)2=(x2+(y+5)2)2\Rightarrow {{\left( \sqrt{{{x}^{2}}+{{\left( y-5 \right)}^{2}}} \right)}^{2}}={{\left( \sqrt{{{x}^{2}}+{{\left( y+5 \right)}^{2}}} \right)}^{2}}
⇒x2+(y−5)2=x2+(y+5)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.
⇒(y−5)2=(y+5)2\Rightarrow {{\left( y-5 \right)}^{2}}={{\left( y+5 \right)}^{2}}
⇒y2−10y+25=y2+10y+25\Rightarrow {{y}^{2}}-10y+25={{y}^{2}}+10y+25
By cancelling out the similar terms from both sides we will get,
⇒−10y=10y\Rightarrow -10y=10y
⇒10y+10y=0\Rightarrow 10y+10y=0
⇒20y=0\Rightarrow 20y=0
Divide the both sides of the equation by 20.
⇒y=020\Rightarrow y=\dfrac{0}{20}
⇒y=0\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.