Exercise 1.14

Question

Exercise 14: If $z \in \mathbb C$, $ |z| = 1$, compute $|1+z|^2 + |1-z|^2$.

ghostofgarborg · Accepted Answer

\begin{align*}
        |1+z|^2 + |1-z|^2 &= (1+z)(\overline{1+z}) + (1-z)(\overline{1-z}) \
        &= (1 + z)(1 + \bar z) + (1 - z)(1 - \bar z) \
        &= (1 + z + \bar z + z \bar z)  + (1 -z - \bar z + z \bar z) \
        &= 2 + 2z \bar z \
        &= 4
    \end{align*}

Amit Awasthi · Answer

\begin{proof}[Computation] Since by Theorem $1.31(a)$ and Definition 1.32,
      \begin{align*}
        |1 + z|^2 &= (1+z)\overline{(1+z)}\
        & = (1+z)(1+\overline{z})\
        &= 1 + z + \overline{z} + z\overline{z} = 2 + z + \overline{z},
        \intertext{and}
        |1 - z|^2 &= (1-z)\overline{(1-z)}\
        &= (1-z)(1-\overline{z})\
        &= 1 - z - \overline{z} + z\overline{z} = 2 - z - \overline{z},
      \end{align*}
      we get
      \begin{equation*}
        |1 + z|^2 + |1 - z|^2 = 4.\qedhere
      \end{equation*}
\end{proof}

Exercise 1.14

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Exercise 1.14

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