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Prove the function $f:\Bbb C^{*} \rightarrow \Bbb R^+$ given by $f(a+bi)=\sqrt{a^2+b^2}$ is a homomorphism and describe its kernel.

Homomorphism:

Let $a,b,c,d$ be in $\Bbb C^{*}$ and for $f$ to be a homomorphism then $f((a+bi)(c+di))=f(a+bi)f(c+di)$

Note how:

$f((a+bi)(c+di))=$

$f((ac-bd)+(bc+ad)i)=$

$\sqrt{(ac-bd)^2+(bc+ad)^2}=$

$\sqrt{(ac)^2+(bd)^2+(ad)^2+(bc)^2}$

and I got stuck here.

Kernel:

Once I am to show that $f$ is a homomorphism then I know there exist a kernel which shall be denoted $k$.

$k=\{a,b \in C^{*} \text { where } f(a+bi)=1\}$

Hopefully someone can point me in the right way to satisfy the property of homomorphism and verify my kernel.

Cameron Williams
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Chilanie
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    Also, I think you mean $z,w\in \mathbb{C}^*$, equivalent to $z=a+bi$ and $w=c+di$ where $a,b,c,d\in \mathbb{R}$ – Eoin Apr 10 '15 at 03:04

2 Answers2

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To finish off the homomorphism proof, note that $f(a+bi)f(c+di)=\sqrt{a^2+b^2}\sqrt{c^2+d^2}$, and then expand under a common radical.

You are correct that this is the kernel. It must map to the identity of the codomain, which in this case is $1$ as $x1=1x=x$ for every $x\in \mathbb{R}^+$. This corresponds to the unit circle of $\mathbb{C}^*$, or all $z\in \mathbb{C}$ such that $z=e^{i\theta}$. You can prove more using Euler's identity.

Eoin
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Once the last line reached , you can factor $\sqrt{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}=\sqrt{a^{2}(c^{2}+d^{2})+b^{2}(c^{2}+d^{2})}=\sqrt{(a^{2}+b^{2})(c^{2}+d^{2})}=f(a+bi)f(c+di)$. The kernel is the subgroup of $\mathbf{C}$ such that $a^{2}+b^{2}=1$ hence the elements of the form $e^{i\theta}$ where $0\leq \theta <2\pi$

mich95
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