Using properties of complex numbers and arguments, proving that statement is the same as proving; arg(z1z2) - arg(z1+z2)^2 = 0 or just that arg(z1z2 / (z1+z2)^2) = 0 whereby now the task is proving that z1z2 / (z1+z2)^2 is a positive real number as long as |z1|=|z2|. Let z1=re^iA and z2=re^iB then z1z2 / (z1+z2)^2 = e^i(A+B) / [e^i2A + 2e^i(A+B) + e^i2B] = 1 / [ e^i(A-B) + e^-i(A-B) + 2 ] = 1 / [ 2cos(A-B) + 2] and this is always a positive quantity since denominator is bounded by 0 and 4.
This problem is broken. The statement isn't true and therefore can't be proved. Disproving is easy with a simple counterexample: Consider z1 = 1+0i and z2 = -1+0i |z1|= sqrt (1^2 + 0^2) = sqrt (1) = 1 |z2|= sqrt [(-1)^2 + 0^2] = sqrt (1) = 1. Therefore |z1|=|z2|, so z1 & z2 satisfy the only given condition. LHS= arg [1*(-1)] = arg (-1) = pi. RHS= arg [(1+0i + (-1+0i) )^2] = arg [(1-1)^2] = arg (0^2) = arg (0), which is undefined. Therefore LHS != RHS QED. In order for the question to be okay, it needs a second condition that z1+z2 != 0.
Using properties of complex numbers and arguments, proving that statement is the same as proving;
arg(z1z2) - arg(z1+z2)^2 = 0
or just that
arg(z1z2 / (z1+z2)^2) = 0
whereby now the task is proving that z1z2 / (z1+z2)^2 is a positive real number as long as |z1|=|z2|.
Let z1=re^iA and z2=re^iB then
z1z2 / (z1+z2)^2 = e^i(A+B) / [e^i2A + 2e^i(A+B) + e^i2B] = 1 / [ e^i(A-B) + e^-i(A-B) + 2 ] = 1 / [ 2cos(A-B) + 2]
and this is always a positive quantity since denominator is bounded by 0 and 4.
Not that it impacts your point, but at the end I believe it should be 1 / [ 2cos(A-B) + 2].
@@JohnSmith-rf1tx good spot thanks :)
Your techniques are stunning. The way you instruct must inspire educators to elevate their strategy.
At 6:40 when expand the square should not be: 2r^2(e^i(a+b)) instead than r^2(e^i(a+b)) ?
Yeah it's a typo I guess
This is what I came here to say.
Exercises vs Problems - nice comparison. Every single Real Analysis question is a problem then (in both senses of the word) 🤣🤣
Are you using obs studio for putting your vdo along with the slides on screen?
Probably not
Nice
This problem is broken. The statement isn't true and therefore can't be proved. Disproving is easy with a simple counterexample:
Consider z1 = 1+0i and z2 = -1+0i
|z1|= sqrt (1^2 + 0^2)
= sqrt (1)
= 1
|z2|= sqrt [(-1)^2 + 0^2]
= sqrt (1)
= 1.
Therefore |z1|=|z2|, so z1 & z2 satisfy the only given condition.
LHS= arg [1*(-1)]
= arg (-1)
= pi.
RHS= arg [(1+0i + (-1+0i) )^2]
= arg [(1-1)^2]
= arg (0^2)
= arg (0), which is undefined.
Therefore LHS != RHS QED.
In order for the question to be okay, it needs a second condition that z1+z2 != 0.
What does it means QED?