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Express the given complex numbers in the form a + bi

by RoRi

Write the following complex numbers in the form a+bi.

  1. (1+i)^2.
  2. \frac{1}{i}.
  3. \frac{1}{1+i}.
  4. (2+3i)(3-4i).
  5. \frac{1+i}{1-2i}.
  6. i^5 + i^{16}.
  7. 1 + i + i^2 + i^3.
  8. \frac{1}{2}(1+i)(1+i^{-8}).
  1. We compute,

        \[ (1+i)^2 = (1+i)(1+i) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i. \]

  2. We compute,

        \[ \frac{1}{i} = \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{-1} = -i. \]

  3. We compute,

        \[ \frac{1}{1+i} = \frac{1-i}{(1+i)(1-i)} = \frac{1-i}{1 - i^2} = \frac{1}{2} - \frac{1}{2}i. \]

  4. We compute,

        \[ (2+3i)(3-4i) = 6 + 9i - 8i -12i^2 = 6 - 12(-1) + i = 18 + i. \]

  5. We compute,

        \[ \frac{1+i}{1-2i} = \frac{(1+i)(1+2i)}{(1-2i)(1+2i)} = \frac{1 + 3i - 2}{1 + 4} = -\frac{1}{5} + \frac{3}{5}i. \]

  6. Since i^4 = 1 we compute as follows

        \[ i^5 + i^{16} = 1 \cdot i + 1^4 = 1 + i. \]

  7. We compute,

        \[ 1 + i + i^2 + i^3 = 1 + i -1 - i = 0. \]

  8. First, since i^4 = 1 we have

        \[ i^{-8} = \frac{1}{i^8} = \frac{1}{1^2} = 1. \]

    Therefore,

        \[ \frac{1}{2}(1+i)(1+i^{-8}) = 1+i. \]

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