S. Verona Lišková
Table of Contents
Exercise 4.1
Easy. Check the division rule for complex numbers
\[\begin{equation*}\frac{a+bi}{c+di} = \frac{ac+bd}{c^2+d^2} + \frac{bc - ad}{c^2 + d^2}i\end{equation*}\]
by multiplying through by \(c + di\) or \(c - di.\)
Solution
Doing so, we have
\[\begin{aligned} a + bi &= (c + di)\left[\frac{ac + bd}{c^2+d^2} + \frac{bc - ad}{c^2 + d^2}i\right] \newline &= \frac{1}{c^2 + d^2}(ac^2 + acdi + bcd + bd^2i + bc^2i + bcdi^2 - acdi - ad^2i^2) \newline &= \frac{1}{c^2 + d^2}(ac^2 + acdi + bcd + bd^2i + bc^2i - bcd - acdi + ad^2) \newline &= \frac{1}{c^2 + d^2}(ac^2 + bd^2i + bc^2i + ad^2) \newline &= \frac{(a + bi)(c^2 + d^2)}{c^2+d^2} = a + bi. \end{aligned}\]
Exercise 4.2
Medium. Check the rules of complex algebra directly from the addition and multiplication rules:
- \(w + z = z + w\);
- \(w + (u + z) = (w + u) + z\);
- \(wz = zw\);
- \(w(uz) = (wu)z\);
- \(w(u + z) = wu + wz\);
- \(w + 0 = w\);
- \(w \cdot 1 = w.\)
Solution
Exercise 4.3
Easy. Let \[z = \sqrt{\dfrac{a + \sqrt{a^2 + b^2}}2} + \sqrt{\dfrac{-a + \sqrt{a^2 + b^2}}2}i\]
Show that \(z^2 = a + bi.\)
Solution
\[\begin{aligned} &\left(\sqrt{\dfrac{a + \sqrt{a^2 + b^2}}2} + \sqrt{\dfrac{-a + \sqrt{a^2 + b^2}}2}i\right)^2 \newline &\qquad= \dfrac{a + \sqrt{a^2 + b^2}}2 + 2\sqrt{\dfrac{a^2 + b^2 - a^2}4}i + \left(\dfrac{-a + \sqrt{a^2 + b^2}}2\right)i^2 \newline &\qquad= \frac{a}2 + \frac{\sqrt{a^2 + b^2}}2 + 2 \left(\frac{b}2\right)i - \left[-\frac{a}2 + \frac{\sqrt{a^2+b^2}}2\right] \newline &\qquad= a + bi. \end{aligned}\]