# show a mapping is homomorphism

Assume a complex number $z=a+bi$ has the length $|z|=\sqrt{a^2+b^2}$, and it has the property that $|z_1z_1|=|z_1||z_2|$. A function $\phi$ from the multiplicative group $\mathbb{C^*}$ of non-zero complex numbers to the additive group $\mathbb{R}$ of real numbers by $\phi(z)=ln(|z|)$.

a). Show that $\phi$ is a group homomorphism

I came up with $\phi(z_1)\phi(z_2)=\phi(z_1+z_2)$ for $\phi$ to be homomorphism. Is this correct? Then I got stuck with $ln(|z_1|)ln(z_2)=ln(|z_1+z_2|)$. I don't know how to show that this equation is true.

b). Describe the Kernel K of $\phi$

Is it true that K=$(z\in G|\phi(x)=0)$ since the identity in the group of all real numbers is zero?

c). Explain why $\mathbb{C^*}/K\approx\mathbb{R}$

Have no clue how to do this one.....

• Let $(G,*_G)$ and $(H,*_H)$ be two groups. A map $f: G\to H$ is a homorophism if $f(g_1*_G g_2)=f(g_1)*_Hf(g_2)$. Your groups are $(\mathbb{C}^*,\cdot)$ and $(\mathbb{R},+)$, your map is $\phi:\mathbb{C}^* \to \mathbb{R}$ defined by $\phi(z)=\ln(|z|)$. Can you find your error? – Nex Nov 20 '15 at 5:52
• Did you mean $|z_1z_2|=|z_1||z_2|$? – Axoren Nov 20 '15 at 7:23

For the first one, you got it backward and it should be $$\varphi(z_1 z_2)=\ln (|z_1 z_2|)=\ln (|z_1||z_2|)=\ln |z_1|+\ln |z_2|=\varphi(z_1)+\varphi(z_2)$$ which shows it is a homomorphism.
Second when is $\ln |z|=0$? when $|z|=1$ which is the unit circle on the complex plane
Third, a wordy explination is that if $a,b\in\mathbb{C}^\ast/\ker \varphi$ and $a=b$ then we have that $ab^{-1}\in\ker\varphi$, which means that $ab^{-1}$ is on the unit circle, when does that happen? when $a=\lambda b$ and $\lambda=\frac{1}{|a|}$, this means that if you view them as vectors that they are scalar multiple of each other and as that is the difference it makes the group in question, but scalars for complex numbers are real numbers, so they are isomorphic.
• Why would $ab^{-1}\in\ker\varphi$ if $a,b∈C∗/kerφ$ – dfdffewfw Nov 20 '15 at 14:09
• My bad, I forgot to add that $a=b$ which I have added now – Zelos Malum Nov 20 '15 at 14:10
• sorry I still don't understand how $ab^{-1}$ is related and is that for showing onto? – dfdffewfw Nov 20 '15 at 14:17