Hw4 Problem 3

Problem 3

In each case below find the inverse of the element under the given operation.
(a) $13$ in $\langle \mathbb{Z}_{20}, +_{20}\rangle$,
(b) $\zeta^{13}$ in $\langle U_{13}, \cdot \rangle$,
(c) $3-2i$ in $\mathbb{C}^*$, the group of nonzero complex numbers under multiplication.

Solution

(a) Identity in $\langle \mathbb{Z}_{20}, +_{20}\rangle$ is $0$. So $13 + _{20}x = 0$ where $x$ is the inverse. So $13 + _{20}7 = 0$, hence, $x=7$.

(b) $\zeta^{13} \cdot \zeta^{0} = \zeta^{13}$ Hence, $\zeta^{0}$ is the inverse.

(c) Under multiplication, the inverse is $1$. So, it is clear the inverse of $3-2i$ is $\frac{1}{3-2i}$. But the number must be in $a+bi$ form. So, $\frac{1}{3-2i} \cdot \frac{3+2i}{3+2i} = \frac{3+2i}{13}$.