Q.The complex number $z$ which satisfies the condition $\left|\dfrac{i+z}{i-z}\right|=1$ lies on:
(A) circle $x^2+y^2=1$
(B) the $x$-axis
(C) the $y$-axis
(D) the line $x+y=1$

The key idea is to use the properties of the modulus of complex numbers and the algebraic representation of a complex number to find its locus.

Given the condition $\left|\dfrac{i+z}{i-z}\right|=1$:

1. Using the property $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$, we can write the condition as $|i+z| = |i-z|$.
2. Let $z = x+iy$. Substitute this into the equation: $|i + (x+iy)| = |i - (x+iy)|$ $|x + i(y+1)| = |-x + i(1-y)|$
3. Apply the definition of modulus, $|a+ib| = \sqrt{a^2+b^2}$, and square both sides to eliminate the square roots: …

The problem asks us to find the locus of a complex number $z$ that satisfies a given condition involving its modulus. Understanding what the modulus of a complex number represents is key here.

The modulus of a complex number $z = x+iy$, denoted $|z|$, is its distance from the origin $(0,0)$ in the Argand plane, calculated as $\sqrt{x^2+y^2}$. More generally, $|z_1 - z_2|$ represents the distance between the complex numbers $z_1$ and $z_2$.

The given condition is $\left|\dfrac{i+z}{i-z}\right|=1$.

We can rewrite $i+z$ as $z - (-i)$ and $i-z$ as $z - i$.

So the condition becomes $\left|\dfrac{z - (-i)}{z - i}\right|=1$.

Using the property that $\left|\dfrac{w_1}{w_2}\right| = \dfrac{|w_1|}{|w_2|}$ for complex numbers $w_1, w_2$ (where $w_2 \neq 0$), we can write:

$\dfrac{|z - (-i)|}{|z - i|} = 1$

This implies $|z - (-i)| = |z - i|$.

This equation states that the distance from $z$ to the complex number $-i$ is equal to the distance from $z$ to the complex number $i$.

In the Argand plane:

• The complex number $i$ corresponds to the point $(0,1)$.
• The complex number $-i$ corresponds to the point $(0,-1)$.

The locus of points that are equidistant from two fixed points is the perpendicular bisector of the line segment joining those two points.

The two fixed points here are $(0,1)$ and $(0,-1)$. The line segment connecting them lies along the $y$-axis. The midpoint of this segment is $(0,0)$. The line that is perpendicular to the $y$-axis and passes through the origin is the $x$-axis.

Therefore, we expect the locus of $z$ to be the $x$-axis. Let's confirm this algebraically.

1. Rewrite the given condition:

   The given condition is $\left|\dfrac{i+z}{i-z}\right|=1$.

   Using the property $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$, we get:

   $\dfrac{|i+z|}{|i-z|} = 1$

2. Equate the moduli:

   Since $|i-z|$ must be non-zero (otherwise $z=i$, which would make the denominator zero), we can multiply both sides by $|i-z|$:

   $|i+z| = |i-z|$

3. Substitute $z = x+iy$:

   Let $z = x+iy$, where $x$ and $y$ are real numbers representing the real and imaginary parts of $z$, respectively.

   Substitute this into the equation:

   • $i+z = i + (x+iy) = x + i(y+1)$
   • $i-z = i - (x+iy) = -x + i(1-y)$

4. Calculate the moduli: …