- #1
sqrt(-1)
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I'm slightly confused with the following function so I was wondering if anybody could give me some hints as to the next step.
A function f is defined as
[tex]
f:\mathbb{C} \longrightarrow \mathbb{C} \\
[/tex]
[tex]
~~z \longmapsto |z|
[/tex]
where
[tex]
\mathbb{C} = (\mathbb{C},+)
[/tex]
assuming the function is a homomorphism, I've gone on to find
[tex]
\rm{Ker}(f) = \{0\}
[/tex]
[tex]
\rm{Im}(f) = \mathbb{R}^+ \cup \{0\}
[/tex]
by the first isomorphism theorem I would have made the following conclusion(this part doesn't make sense to me):
[tex]
\mathbb{C}/\rm{Ker}(f) \cong (\mathbb{R}^+ \cup \{0\}, + )
[/tex]
But firstly
[tex]
(\mathbb{R}^+ \cup \{0\}, + )
[/tex]
is not a group since there is no inverse element, and secondly since Ker(f) = the trivial group
[tex]
\mathbb{C}/\rm{Ker}(f) \cong (\mathbb{C}, + )
[/tex]
Could somebody give me some pointers please?
A function f is defined as
[tex]
f:\mathbb{C} \longrightarrow \mathbb{C} \\
[/tex]
[tex]
~~z \longmapsto |z|
[/tex]
where
[tex]
\mathbb{C} = (\mathbb{C},+)
[/tex]
assuming the function is a homomorphism, I've gone on to find
[tex]
\rm{Ker}(f) = \{0\}
[/tex]
[tex]
\rm{Im}(f) = \mathbb{R}^+ \cup \{0\}
[/tex]
by the first isomorphism theorem I would have made the following conclusion(this part doesn't make sense to me):
[tex]
\mathbb{C}/\rm{Ker}(f) \cong (\mathbb{R}^+ \cup \{0\}, + )
[/tex]
But firstly
[tex]
(\mathbb{R}^+ \cup \{0\}, + )
[/tex]
is not a group since there is no inverse element, and secondly since Ker(f) = the trivial group
[tex]
\mathbb{C}/\rm{Ker}(f) \cong (\mathbb{C}, + )
[/tex]
Could somebody give me some pointers please?
Last edited: