Question

The number of complex numbers z such that $\left| z+i \right|=\left| z-1 \right|=\left| z+1 \right|$
$\begin{align}
  & \text{a) 0} \\
 & \text{b) 1} \\
 & \text{c) 2} \\
 & \text{d) None of these} \\
\end{align}$

Answer 1

Hint: Now to solve the equation we will consider three equations $\left| z+1 \right|=\left| z-1 \right|$, $\left| z+1 \right|=\left| z+i \right|$ and $\left| z+i \right|=\left| z-1 \right|$. Now in this equation we will substitute z as $x+iy$ and simplify the equations to find the values of x and y. Hence we will get the number of solutions of the equation.

Complete step by step solution:
Now consider the given equation $\left| z+i \right|=\left| z-1 \right|=\left| z+1 \right|$.
Let us say $z=x+iy$
Now first consider the equation $\left| z-1 \right|=\left| z+1 \right|$
Now substituting the equation $z=x+iy$ we get,
$\begin{align}
  & \Rightarrow \left| x+iy-1 \right|=\left| x+iy+1 \right| \\
 & \Rightarrow \left| \left( x-1 \right)+iy \right|=\left| \left( x+1 \right)+iy \right| \\
\end{align}$
Now we know that $\left| a+ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$
$\Rightarrow \sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}}=\sqrt{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}$
Now squaring both sides of the equation we get,
$\Rightarrow {{\left( x-1 \right)}^{2}}+{{y}^{2}}={{\left( x+1 \right)}^{2}}+{{y}^{2}}$
Subtracting ${{y}^{2}}$ on both sides of the equation we get,
$\Rightarrow {{\left( x-1 \right)}^{2}}={{\left( x+1 \right)}^{2}}$
Now let us open the bracket by using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ Hence we get,
$\Rightarrow {{x}^{2}}-2x+1={{x}^{2}}+2x+1$
Now we will again subtract ${{x}^{2}}+1$ on both sides we get,
$\Rightarrow -2x=2x$
Now the only solution to this is x = 0.
Hence we get x = 0.
Now consider the equation $\left| z+i \right|=\left| z-1 \right|$
Now substituting the value of z we get,
$\begin{align}
  & \Rightarrow \left| x+iy+i \right|=\left| x+iy-1 \right| \\
 & \Rightarrow \left| x+\left( y+1 \right)i \right|=\left| \left( x-1 \right)+iy \right| \\
\end{align}$
Now again using the definition of modulus of complex numbers we get,
$\Rightarrow \sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}=\sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}}$
Now squaring the number on both sides we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( x-1 \right)}^{2}}+{{y}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+2y+1={{x}^{2}}+1-2x+{{y}^{2}} \\
 & \Rightarrow 2y=-2x \\
\end{align}$
Now substituting the value of x in the equation we get y = 0
Hence we have x = y = 0.
Now consider the equation $\left| z+i \right|=\left| z+1\right|$
Now let us simplify this equation too in the same manner. Hence we get,
$\begin{align}
  & \Rightarrow \left| x+iy+i \right|=\left| x+iy+1 \right| \\
 & \Rightarrow \left| x+\left( y+1 \right)i \right|=\left| \left( x+1 \right)+iy \right| \\
 & \Rightarrow {{x}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( x+1 \right)}^{2}}+{{y}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+2y+1={{x}^{2}}+1+2x+{{y}^{2}} \\
 & \Rightarrow 2y=2x \\
 & \Rightarrow x=y \\
\end{align}$
Hence the only solution of this equation x = y = 0.

So, the correct answer is “Option b”.

Note: Now note that the modulus of complex numbers is nothing but the distance of the complex number from origin in the complex plane. Hence we can calculate this distance using the Pythagoras theorem.