Log(z_1 z_2) = log_z1 + log_z2 + 2N(pi)r ~ How do I calculate Arg(z)? Stuck on complex variables...?

Question

I'm stuck on this question in my text book

Show for any 2 non-zero complex numbers

Log(z_1 z_2) = log_z1 + log_z2 + 2N(pi)r

Where N is either 0, +1, -1

I'm pretty good normally with this stuff, but the notes (or text) don't cover how to calc arg(z) or really explain the whole chapter very well

This is what I know

Log(z_1 z_2) = log_z1 + log_z2 (this is a formula that is given - so my main prob is that is the question and answer with out the 2N(pi)r... so how does it also equal the quesiton I'm asking???)

I also know

log(z) = ln|z| + iArg|z| + or = 2 (pi)ki (where k=0,1,2..)

and log(z) = ln + i(\omega + 2n(pi))

I can almost see the answers in the formulas but don't understand how they work because they all claim to equal log(z) so which ones to I use in the product to get the answer, and also at what point do I need to use the values for N ?

Any help or advice will be much appreciated! :)

Answer 1

log(z_1*z_2) = log( r_1 * e^(i * T_1) * r_2 * e^(i * T_2) )
= log( r_1 * r_2 * e^(i * (T_1 + T_2)) )
= ln(r_1 * r_2) + i * ((T_1 + T_2) + 2 * k * Pi)
= ln(r_1) + i * T_1 + ln(r_2) + i * T_2 + i * 2 * k * Pi
= log(z_1) + log(z_2) + i * 2 * k * Pi

The above is (unnecessarily) elaborate to show the different steps (hopefully) more clearly.

Note that when z is written as : z = r * e^(i * T) with r > 0 and -Pi < T <= Pi , we get :

log(z) = ln(r) + i * (T + 2 * k * Pi)

for any integer k.

Answer 2

i did not understand what you were asking but it is really log(z1z2) = logz1 + log z2 for all z1, z2 greater than 0. but i think log(z1z2) = logz1 + logz2 + 2n(pi)r for all nonzero complex numbers. example: log 1 = log1 + log1, 1 =1cis90 where r =1.
log1 = logz1 + logz2 + 2(0)pi(1) = log1 + log1

Answer 3

DO YOUR OWN HOME WORK lol