MHB Is It Possible to Prove the Complex Number Challenge?

[Greg]

Prove that $\arg[(a+bi)(c+di)]=\arg(a+bi)+\arg(c+di)$.

 

[kaliprasad]

Prove that $\arg[(a+bi)(c+di)]=\arg(a+bi)+\arg(c+di)$.

Spoiler

we have $(a+bi)(c+di) = (ac - db) + (bc + ad)i$
hence $\arg[(a+bi)(c+di)] = \arg (ac - db) + (bc + ad)i$
or $\arg[(a+bi)(c+di)] = \arctan (\frac{bc + ad}{ac-db})$
$= \arctan (\frac{\frac{b}{a}+ \frac{d}{c}}{1 - \frac{b}{a}\frac{d}{c}})$
$= \arctan (\tan ( \arctan(\frac{b}{a}) + \arctan(\frac{d}{c}))$
$= \arctan(\frac{b}{a}) + \arctan(\frac{d}{c})$
$=\arg(a+bi)+\arg(c+di)$.

 

[Nono713]

Spoiler

Let $z_1 = a + bi$ and $z_2 = c + di$. Express these complex numbers in polar form as:
$$z_1 = r_1 e^{i \theta_1}$$
$$z_2 = r_2 e^{i \theta_2}$$
By simple complex multiplication:
$$z_1 z_2 = r_1 e^{i \theta_1} r_2 e^{i \theta_2} = r_1 r_2 e^{i \left ( \theta_1 + \theta_2 \right )}$$
It follows by definition of the complex argument that:
$$\mathrm{arg} ~ z_1 = \theta_1$$
$$\mathrm{arg} ~ z_2 = \theta_2$$
$$\mathrm{arg} ~ z_1 z_2 = \theta_1 + \theta_2 = \left ( \mathrm{arg} ~ z_1 \right ) + \left ( \mathrm{arg} ~ z_2 \right )$$
$$\blacksquare$$

 

[alyafey22]

Prove that $\arg[(a+bi)(c+di)]=\arg(a+bi)+\arg(c+di)$.

I suppose you mean the principle argument because the given function is multivalued!

 

[Greg]

I suppose you mean the principle argument because the given function is multivalued!

Yes, thanks.

 

[Deveno]

I note, in passing, that the proposition stated cannot hold if just one of the two complex numbers is zero (even if one adopts the convention that $\arg(0) = 0$).

 

[Opalg]

I suppose you mean the principle argument because the given function is multivalued!

On the contrary, the result is false if you insist on the principal value of the argument, because the left and right sides can differ by $2\pi.$

 

[alyafey22]

On the contrary, the result is false if you insist on the principal value of the argument, because the left and right sides can differ by $2\pi.$

I don't see how is that possible ? Give an example please.

 

[alyafey22]

On the contrary, the result is false if you insist on the principal value of the argument, because the left and right sides can differ by $2\pi.$

OK, I think the values should be taking modulo $2\pi$.

 

[Deveno]

The situation is this:

The complex exponential turns sums of angles into products of unit complex numbers. But this function is not injective, so taking the inverse (which is essentially what this problem asks us to do) requires some qualification.

 

[Greg]

I should have put more thought into what I was asking. My apologies and thanks for the replies.

 

[alyafey22]

I should have put more thought into what I was asking. My apologies and thanks for the replies.

Such topics are confusing sometimes and an extra care must be taking when dealing with them. I personally face difficulties dealing with some of these concepts.