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- เผยแพร่เมื่อ 7 ก.พ. 2025
- ▶ Greetings, everyone! Welcome to @aplusbi 🧡🤩💗
This channel is dedicated to the fascinating realm of Complex Numbers. I trust you'll find the content I'm about to share quite enjoyable. My initial plan is to kick things off with informative lectures on Complex Numbers, followed by a diverse range of problem-solving videos.
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Cool!
Nice! From all of your channels aplusbi is my favorite😊😊💯💯💯💥💥💥💥💥💪💪💪💪
Thanks 🔥
I wonder if it is allowed to multiply both exponents by 1/n , then try to find solutions for m/n
Why not? Check the comment from @Don-Ensley
problem
( 1 + i ) ᵐ = ( 1 - i ) ⁿ
First obvious solution is m = n = 0.
Given n 0, raise to the power 1/n on both sides.
( 1 + i ) ᵐᐟⁿ= ( 1 - i )
Take natural logarithms on both sides. Bring down exponents.
(m/n) ln (1+i) = ln( 1-i)
(m/n) π(1/4 +2 k) = -π (1/4+2p)
(m/n) (1 +8 k) = - (1+8p)
(m/n) (8 k + 1) = - (8p + 1)
(8 k + 1) m = - (8p + 1) n
(m/n) = - (8p + 1) / (8 k + 1)
k, p ∈ ℤ
answer
(m,n) ∈ { (0,0), (- (8p + 1), (8 k + 1)),
(k, p ∈ ℤ ) }
Here's a geometric version. Think of 1+i and 1-i as hands on the clock, pointing at +/- 45 deg.
One 'tick' is one power increase - a movement of +/- 45 deg, with a magnitude increase of (^sqrt2.)
When do the 2 hands meat? 0 ticks at 0 deg, then 4 ticks at 180, 8 when they pass ea other again at 0 deg, etc. m = n = 4k, k >= 0.
Wow, that’s awesome! 🤩
e^(-i*k*pi)=e(i*k*pi) when n is even
k is even :)
m=n=4k I believe. I'll check with the video answer in a minute.
The first equality is because the base of both sides has modulus sqrt2, and therefore the end result of each side has modulus 2^(m/2)=2^(n/2).
Once we know m=n, we have a limited search space, and we only need to consider direction. The bases of the two sides have angles +/- pi/4, rotating this by pi/4 with each power, and the two nth powers meet up exactly when n is a multiple of 4.
Ok I did not look for complex solutions, that's on me. 😂
To be fair, there are weird considerations about branches and definitions for nonreal base and noninterger exponents. 😅
😄