The locus represented by $\left| z-1 \right|=\left| z+i \right|$ is
A) circle of radius 1 unit
B) An ellipse with foci at (1,0) and (0,1)
C) A straight line through the origin
D) A circle the line joining (1,0) and (0,1) as diameter

Hint: We need to know the concept of complex number to solve the given question. The complex number $z$ is written as $x+iy$. The further calculation is done considering $x$ and $y$as $z$. We need to find the value of modulus of the complex function, so that we can find the relation between $x$ and $y$. As per the relation we can find what does the equation represents.

Complete step by step solution:
The question here is in complex number represented as $z$. The question ask us to find the locus of $\left| z-1 \right|=\left| z+i \right|$ .
Consider the complex number $z$as $x,y$ mathematically $z=x+iy$ , where $x$ and $y$ are the real numbers. So we can represent $\left| z-1 \right|$ as$\left| x+iy-1 \right|$ and $\left| z+i \right|$ as $\left| x+iy+i \right|$ . Writing it with equal to sign we get,
Taking the function in L.H.S
$\left| z-1 \right|=\left| x+iy-1 \right|$
Taking real numbers and complex number inside the modulus we get:
$\Rightarrow \left| z-1 \right|=\left| \left( x-1 \right)+iy \right|$
Taking the function in R.H.S
$\left| z+i \right|=\left| x+iy+i \right|$
Similarly, taking the real and complex number together
$\Rightarrow \left| z+i \right|=\left| x+i\left( y+1 \right) \right|$
After the above calculation \[\left| z-1 \right|=\left| z+i \right|\] could be written as:
\[\Rightarrow \left| x-1+iy \right|=\left| x+i\left( y+1 \right) \right|\]
Modulus of $z$,$\left| z \right|$means $\sqrt{{{x}^{2}}+{{y}^{2}}}$, which means the sum of the square of the real and irrational number so applying the same with the function we achieved in the equation we get:
\[\Rightarrow \sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}}=\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}\]
Squaring both side to find the make the calculation easier, we get:
\[\Rightarrow {{\left( x-1 \right)}^{2}}+{{y}^{2}}={{x}^{2}}+{{\left( y+1 \right)}^{2}}\]
On expanding the expression given, we get:
\[\Rightarrow {{x}^{2}}+1-2x+{{y}^{2}}={{x}^{2}}+{{y}^{2}}+1+2y\]
On calculating further we get:
\[\Rightarrow {{x}^{2}}+1-2x+{{y}^{2}}-{{x}^{2}}-{{y}^{2}}-1-2y=0\]
\[\Rightarrow -2x-2y=0\]
Multiplying $-\dfrac{1}{2}$ to both the terms in L.H.S and R.H.S, we get:
$\Rightarrow x+y=0$
$\therefore $ The locus represented by $\left| z-1 \right|=\left| z+i \right|$ is $(C)$ straight line through origin.

So, the correct answer is “Option C”.

Note: The equations of the straight line in the answer do not have any constant which means the straight line passes through the origin. The complex number is represented as a sum of real numbers to make the calculation easy. Different relations between complex numbers form different locus representing different 2-dimensional figures.