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Abstract Algebra

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Moderate

Graph of a Composition of Functions

ABSALG-EIMLM1

One definition of a function is like associating to a function its graph.

We define a function $f$ from $S$ to $T$ as a subset $R_f$ of the Cartesian product $S\times T$ such that $(s,t_1)$ and $(s,t_2)$ are in $R_f$ if and only if $t_1=t_2=f(s)$.

In this description, if we have $h=(g\circ f)$, then $R_f\subseteq S\times T$, $R_g\subseteq T\times U$ and $R_h\subseteq S\times U$.

Let $S=\{1,i\}$, $T=\{-\pi<\theta\le \pi\}$ and $U=\{1,i,-1,-i\}$, where $i=\sqrt{-1}$.

Which of the following gives a correct statement about $R_h$ for the given $R_f$, $R_g$?

A

With $R_f=\{(s,\theta)\mid e^{i\theta}=s^4\}$, $R_g=\{(\theta,u)\mid u=e^{i\theta/3}\}$ we have:

$$R_h=\{(1,e^{2\pi i/3}), (i,e^{2\pi i/3})\}$$

B

With $R_f=\{(s,\theta)\mid e^{i\theta}=s^4\}$, $R_g=\{(\theta,u)\mid u=e^{i\theta/3}\}$ we have:

$$R_h=\{(1,1),(1,e^{2\pi i/3}), (1,e^{-2\pi i/3}), (i,1),(i,e^{2\pi i/3}), (i,e^{-2\pi i/3})\}$$

and $R_h$ does not represent a function.

C

With $R_f=\{(s,\theta)\mid s\in S\;{\rm and}\; \theta\in T\; {\rm with} \; e^{3i\theta}=s^4\}$, $R_g=\{(\theta,e^{2i\theta})\}$, we have:

$$R_h=\{(1,i), (i,i)\}$$

D

With $R_f=\{(s,\theta)\mid e^{i\theta}=s\}$, $R_g=\{(\theta,u)\mid\,\, u^3=e^{i\theta}\}$ we have:

$$R_h=\{(1,1),(1,e^{2\pi i/3}), (1,e^{-2\pi i/3}), (i,e^{\pi i/6}), (i,e^{4\pi i/6}), (i, e^{-3\pi i/6})\}$$

and $R_h$ does not represent a function.

E

With $R_f=\{(s,\theta)\mid s=e^{i\theta}\}$, $R_g=\{(\theta,u)\mid\,{\rm with}\, u^3=e^{i\theta}\}$ we have:

$$R_h=\{(1,1), (i,-i)\}$$