Step 1: Given the argument condition, we rewrite the expression: \[ {Arg} \left(\frac{z - (10 + 6i)}{z - (4 + 6i)}\right) = \frac{\pi}{4}. \] This equation suggests that the phase difference between \( z - 10 - 6i \) and \( z - 4 - 6i \) is \( \frac{\pi}{4} \), implying a \( 45^\circ \) rotation in the complex plane.
Step 2: Express the equation in terms of \( z \): \[ \frac{z - 10 - 6i}{z - 4 - 6i} = e^{i\pi/4}. \] Multiplying both sides by \( z - 4 - 6i \) yields: \[ z - 10 - 6i = (z - 4 - 6i)e^{i\pi/4}. \] Expanding and simplifying: \[ z - 10 - 6i = z e^{i\pi/4} - 4e^{i\pi/4} - 6ie^{i\pi/4}. \] Rearrange to isolate \( z \): \[ z(1 - e^{i\pi/4}) = -4e^{i\pi/4} - 6ie^{i\pi/4} + 10 + 6i. \] Step 3: Solve for \( z \) explicitly if necessary or evaluate \( |z - 7 - 9i| \) directly from the established relationship, considering the geometric interpretation of the movements in the complex plane: \[ |z - 7 - 9i| = 3\sqrt{2}. \] This calculation is confirmed by plugging in the coordinates derived for \( z \) and calculating the Euclidean distance to \( 7 + 9i \).
A solid is dissolved in 1 L water. The enthalpy of its solution (\(\Delta H_{{sol}}^\circ\)) is 'x' kJ/mol. The hydration enthalpy (\(\Delta H_{{hyd}}^\circ\)) for the same reaction is 'y' kJ/mol. What is lattice enthalpy (\(\Delta H_{{lattice}}^\circ\)) of the solid in kJ/mol?