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Find all real numbers satisfying the given complex equation

For each of the following expressions find all x,y \in \mathbb{R} such that the formula holds.

  1. x + iy = xe^{iy}.
  2. x + iy = ye^{ix}.
  3. e^{x+iy} = -1.
  4. \displaystyle{\frac{1+i}{1-i} = xe^{iy}}.

  1. We use the definition of the complex exponential we compute,

        \[  x + iy = xe^{iy} \quad \implies \quad x + iy &= x \cos y + x (i \sin y). \]

    Then we set the real parts and the imaginary parts on each side equal to each other,

        \[ x = x \cos y \quad \text{and} \quad y = x \sin y. \]

    This implies y = 0 and x is arbitrary. (Since x = x \cos y implies \cos y = 1 which is only true for y = 2n \pi. But then, \sin (2n \pi) = 0 for all n; hence, we must have y = 0 from the second equation. If y = 0 then both equations hold for all x.)

  2. Similarly to part (b) we have,

        \[ x + iy = ye^{ix} \quad \implies \quad x + iy = y \cos x + i (y \sin x). \]

    Setting real and imaginary parts equal we have,

        \[ y \cos x = x, \quad \text{and} \quad y \sin x = y. \]

    This implies x = y = 0.

  3. Again, we compute

        \[ e^{x+iy} = -1 \quad \implies \quad e^x \cos y + i (e^x \sin y) = -1. \]

    Therefore,

        \[ e^x \cos y = -1, \quad \text{and} \quad e^x \sin y = 0. \]

    This implies

        \[ \sin y = 0 \quad \implies quad y = n \pi. \]

    But, this implies x = 0 and so y = (2n+1)\pi.

  4. Finally,

        \[ \frac{1+i}{1-i} = xe^{iy} \quad \implies \quad i = x \cos y + i (x \sin y). \]

    Setting real and imaginary parts equal we have

        \[ x \cos y = 0, \quad \text{and} \quad x \sin y = 1. \]

    This implies

        \[ x = 1, \quad y = \frac{\pi}{2} + 2n \pi, \qquad n \in \mathbb{Z}. \]

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