We are given the equation: \[ \frac{|z - 5i|}{|z - 5i|} = 1 \] The expression \( \frac{|z - 5i|}{|z - 5i|} \) represents the ratio of the magnitude of \( z - 5i \) to itself. This ratio is always 1 unless \( |z - 5i| = 0 \), in which case the ratio would be undefined. Thus, the condition \( \frac{|z - 5i|}{|z - 5i|} = 1 \) implies that \( |z - 5i| \neq 0 \), or equivalently: \[ z \neq 5i. \] This means the point \( z \) cannot be at \( 5i \) on the imaginary axis. Now, we consider the nature of \( z \). Let \( z = x + iy \), where \( x = {Re}(z) \) is the real part and \( y = {Im}(z) \) is the imaginary part of \( z \). The expression \( |z - 5i| \) represents the distance between the complex number \( z = x + iy \) and the point \( 5i \), which is \( (0, 5) \) on the imaginary axis. The distance formula gives: \[ |z - 5i| = \sqrt{x^2 + (y - 5)^2}. \] For the ratio \( \frac{|z - 5i|}{|z - 5i|} = 1 \) to hold, the complex number \( z \) must be such that the imaginary part \( y \) must be equal to zero because if the imaginary part were non-zero, the expression would not yield a ratio of 1.
Hence, the condition simplifies to: \[ {Im}(z) = 0. \]
Therefore, the imaginary part of \( z \) must be zero, which corresponds to option (E).
Thus, the correct answer is \( \boxed{{Im}(z) = 0} \), corresponding to option (E).
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to: