Determine all real numbers satisfying given relations

Determine all values for the real numbers $x$ and $y$ such that the following equations hold.

$x + iy = x - iy$ .
$x + iy = |x + iy|$ .
$|x+iy| = |x-iy|$ .
$(x+iy)^2 = (x-iy)^2$ .
$\frac{x+iy}{x-iy} = x - iy$ .
$\sum_{k=0}^{100} i^k = x + iy$ .

The equation
$x+iy = x-iy \quad \implies \quad 2iy = 0 \quad \implies \quad y = 0.$

The value of $x$ is arbitrary.
Using the formula for the absolute value of a complex number we have
$\begin{align*} x+iy = |x+iy| && \implies && x+iy = \sqrt{x^2 + y^2} \\ && \implies && x = \sqrt{x^2+y^2} \text{ and } y = 0. \end{align*}$

Since $y = 0$ the equation $x = \sqrt{x^2+y^2}$ implies $x = \sqrt{x^2}$ which implies $x \geq 0$ . Therefore, this equation is satisfied by

$x \geq 0, \qquad y = 0.$

(Note: The answer Apostol gives says $x > 0$ , but I think $x = 0$ works as well.
Again, using the formula for the absolute value of a complex number we have
$\begin{align*} |x + iy| = |x-iy|&& \implies && \sqrt{x^2 + y^2} &= \sqrt{x^2 + (-y)^2} \\ && \implies && \sqrt{x^2+y^2} &= \sqrt{x^2 + y^2}. \end{align*}$

This holds for all real $x$ and $y$ .
We compute as follows,
$\begin{align*} (x+iy)^2 = (x-iy)^2 && \implies && x^2 - y^2 + (2xy)i &= x^2 - y^2 - (2xy)i \\ && \implies && (4xy)i = 0. \end{align*}$

Hence, we must have either $x = 0$ and $y$ is arbitrary or $x$ arbitrary and $y =0$ .
We compute,
$\begin{align*} \frac{x+iy}{x-iy} = x-iy && \implies && x+iy &= (x-iy)^2 \\ && \implies && x+iy &= x^2 - y^2 - (2xy)i. \end{align*}$

This gives us two equations (since the real parts and imaginary parts must be equal),

$x = x^2 - y^2, \qquad y = -2xy.$

If $y \neq 0$ then from the second equation we have

$x = -\frac{1}{2} \quad \implies \quad -\frac{1}{2} = \left( -\frac{1}{2} \right)^2 - y^2 \quad \implies \quad y = \frac{\sqrt{3}}{2}.$

If $y = 0$ then we have $x = x^2$ so $x = 0$ or $x = 1$ . But, $x = 0$ is not impossible since then $\frac{x+iy}{x-iy}$ is undefined. Therefore we have two possibilities

$x = 1, \ y = 0 \qquad \text{or} \qquad x = -\frac{1}{2}, \ y = \frac{sqrt{3}}{2}.$

(Note: Apostol only gives the first of these solutions. We can check by a direct substitution that the second solution also works.)
Here we note that
$i^k = \begin{cases} 1 &\text{if } k \equiv 0 \mod 4 \\ i & \text{if } k \equiv 1 \mod 4 \\ -1 & \text{if } k \equiv 2 \mod 4 \\ -i & \text{if } k \equiv 3 \mod 4. \end{cases}$

Therefore, we have

$\begin{align*} \sum_{k=0}^{100} i^k &= \sum_{k=0}^{25} i^{4k} + \sum_{k=0}^{24} i^{4k+1} + \sum_{k=0}^{24} i^{4k+2} + \sum_{k=0}^{24} i^{4k+3} \\ &= \sum_{k=0}^{25} 1 + \sum_{k=0}^{24} i + \sum_{k=0}^{24} (-1) + \sum_{k=0}^{24} (-i) \\ &= 26 + 25i - 25 - 25i \\ &= 1 \end{align*}$

Therefore, from the given equation we have

$x + iy =1 \quad \implies \quad x = 1, \ \ y = 0.$