Question

Find the number of complex numbers \[z\] such that \[\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right|\] is
A. \[0\]
B. \[1\]
C. \[2\]
D. Infinity

Answer 1

Hint: We have to solve this question using the complex number \[z = x + iy\] form then we will use the concept of modulus of a complex number.

Formula used:
If \[z = x + iy\], the modulus of \[z\] is \[\left| z \right| = \sqrt {{x^2} + {y^2}} \].

Complete step by step solution:
 \[\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right|\]
Now let's take the complex number general form as \[z = x + iy\] and we will start substituting the values of mod in the general form .
\[\left| {z - 1} \right| = \left| {(x - 1) + iy} \right| \Rightarrow \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} \] --------- (i)
Same we will be following with \[\left| {z + 1} \right|\] and \[\left| {z - i} \right|\] ,
\[\left| {z + 1} \right| = \left| {(x + 1) + iy} \right| \Rightarrow \sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} \] --------- (ii)
\[\left| {z - i} \right| = \left| {x + i\left( {y - 1} \right)} \right| \Rightarrow \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} \] --------- (iii)

Now we are considering equation (i) is equal to equal (ii) as given in the question .
\[\left| {z - 1} \right| = \left| {z + 1} \right|\]
\[\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} \] = \[\sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} \]
Squaring on both sides.
\[{\left( {x - 1} \right)^2} + {y^2}\] = \[{\left( {x + 1} \right)^2} + {y^2}\]
\[{x^2} + 1 - 2x + {y^2} = {x^2} + 1 + 2x + {y^2}\]
\[4x = 0\]
Hence , \[x = 0\]

Now we are considering equation (i) is equal to equal (iii) as given in the question.
\[\left| {z - 1} \right| = \left| {z - i} \right|\]
\[\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} \]\[ = \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} \]
Now, Put the value of \[x = 0\]
\[\sqrt {\left( {{{\left( { - 1} \right)}^2} + {y^2}} \right)} = \sqrt {{{\left( {y - 1} \right)}^2}} \]
\[1 + {y^2} = {y^2} + 1 - 2y\]
\[2y = 0\]
\[y = 0\]
Hence we have found the values for \[x\] and \[y\] . where \[x = 0\] and \[y = 0\].
\[z = \left( {0,0} \right) = 0 + 0i\]
Hence,
Only One Value of \[z\] is possible i.e \[0 + 0i\] .
Hence the answer is (B) which is \[1\] .

Note: In this solution we have just used the basic concept of complex number where a student has to remember the fundamental equation of complex numbers and after using the equation we have just substituted the values and as per question we have equated the mods. This is the easiest method to solve the question.