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Determine all real numbers satisfying given relations

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

  1. x + iy = x - iy.
  2. x + iy = |x + iy|.
  3. |x+iy| = |x-iy|.
  4. (x+iy)^2 = (x-iy)^2.
  5. \frac{x+iy}{x-iy} = x - iy.
  6. \sum_{k=0}^{100} i^k = x + iy.

  1. The equation

        \[ x+iy = x-iy \quad \implies \quad 2iy = 0 \quad \implies \quad y = 0. \]

    The value of x is arbitrary.

  2. 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.

  3. 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.

  4. 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.

  5. 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.)

  6. 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

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

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