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Sketch inequalities in the complex plane

Sketch each of the following sets of complex numbers z that satisfy the given inequalities:

  1. |2z + 3| < 1.
  2. |z+1| < |z-1|.
  3. |z-i| \leq |z+i|.
  4. |z| \leq |2z+1|.

  1. Letting z = x+ iy we have,

        \begin{align*}  |2z+3| < 1 && \implies && |2x+3 + 2yi| &< 1 \\  && \implies && \sqrt{(2x+3)^2 + 4y^2} &< 1 \\  && \implies && (2x+3)^2 + 4y^2 < 1. \end{align*}

    This is a disk of radius \frac{1}{2} centered at \left(-\frac{3}{2}, 0 \right). The sketch is as follows:
    Complex4

  2. Letting z = x+ iy we have,

        \begin{align*}  |z+1| < |z-1| && \implies && \sqrt{(x+1)^2 +y^2} &< \sqrt{(x-1)^2 + y^2} \\  && \implies && x^2 + 2x + 1 + y^2 &< x^2 -2x + 1 + y^2 \\  && \implies && 4x &< 0 \\  && \implies && x &< 0. \end{align*}

    This is the half-plane with negative real part. The sketch is as follows:
    Complex5

  3. Letting z = x + iy we have,

        \begin{align*}  |z-i| \leq |z+i| && \implies && \sqrt{x^2 + (y-1)^2} &\leq \sqrt{x^2 + (y+1)^2} \\  && \implies && x^2 + y^2 - 2y + 1 &\leq x^2 + y^2 + 2y + 1 \\  && \implies && y & \geq 0. \end{align*}

    This is the half-plane with positive imaginary part. The sketch is as follows:
    Complex6

  4. Letting z = x+ iy we have,

        \begin{align*}  |z| \leq |2z+1| && \implies && \sqrt{x^2+y^2} &\leq \sqrt{(2x+1)^2 + (2y)^2} \\  && \implies && x^2 + y^2 &\leq 4x^2 + 4x + 1 + 4y^2 \\  && \implies && 0 &\leq 3x^2 + 4x + 1 + 3y^2 \\  && \implies && 0 &\leq x^2 + \frac{4}{3}x + \frac{4}{9} + y^2 - \frac{1}{9} \\  && \implies && \frac{1}{9} &\leq \left( x +\frac{2}{3} \right)^2 + y^2. \end{align*}

    This is the region outside the disk of radius \frac{1}{3} centered at the point \left( -\frac{2}{3}, 0 \right). The sketch is as follows:
    Complex7

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