The number of complex numbers \(z\) such that \(|z-1|\) \(=|z+1|=|z-i|\) equal (a) 0 (b) 1 (c) 2 (d) \(\infty\)

Let's first understand the geometric interpretation of the given conditions. The complex number \(z\) can be represented in the complex plane as a point. The distance between two complex numbers \(z_1\) and \(z_2\) is given by the magnitude of their difference, i.e., \(|z_1-z_2|\). So, the condition \(|z-1|=|z+1|\) means that the distance between \(z\) and \(1\) is equal to the distance between \(z\) and \(-1\). This implies that \(z\) lies on the perpendicular bisector of the line segment joining \(1\) and \(-1\) in the complex plane, which is just the imaginary axis. Similarly, the condition \(|z-1|=|z-i|\) implies that \(z\) lies on the perpendicular bisector of the line segment joining \(1\) and \(i\). Now, we need to find the intersection points of the imaginary axis and the other perpendicular bisector.

The imaginary axis in the complex plane contains complex numbers of the form \(zi\), where \(z = a+bi\) and \(a = 0\). Since \(z\) lies on the imaginary axis, let's consider \(z=bi\) for some real number \(b\). Then, given the condition \(|z-1|=|z-i|\), substituting \(z=bi\), we obtain: \[|bi-1|=|bi+i\]

To find the intersection points, let's solve the equation from the previous step. We have \[|bi-1|=|bi+i\] Squaring both sides (since distances are non-negative, this will not introduce extraneous solutions), we get: \[(b^2+(-1)^2)=((b+1)^2+1^2)\] Simplifying the equation, and solving for \(b\): \[b^2+1=b^2+2b+1+1\] \[2b+1=1\] \[b=\frac{1}{2}\] Thus, the intersection point is \(z=\frac{1}{2}i\). #Conclusion# We found that there is only one complex number \(z\) that satisfies the given conditions: \(z=\frac{1}{2}i\). Thus, the correct answer is (b) 1.