A Nice Rational Equation | Problem 467

[@scottleung9587]

I used the second method.

[@shmuelzehavi4940]

2 remarks about the 2'nd method (which is much more elegant than the 1'st one): 1. It's easy to show that: 1 + i = i (1 - i) and thus: (1 + i) / (1 - i) = i 2. For completeness you have to prove that there is no additional complex z≠4n such that: i^z = 1.

[@RexxSchneider]

I'm not sure that anyone will believe that i^z = 1 has other solutions, but for completeness: i^z = exp(2nπi) | n ∈ ℤ. Taking natural logs: z*ln(i) = 2nπi. Giving: z*πi/2 = 2nπi. So z = 4n. Of course, one could then suggest that you have to show that exp(2nπi) is the only representation of 1 in polar form, but I think most folk will accept that. Otherwise you have use Euler and rely on arccos(1) = 2nπ, but where does all that end? We have to assume some knowledge of complex numbers, surely?

[@alikazemi1329]

👏👏👏👏

[@RexxSchneider]

Before watching the video: (1+i)/(1-i) = ( (1+i)(1+i) ) / ( (1-i)(1+i) ) = ( 1 - 1 + 2i ) / ( 1 - (-1) ) = 2i / 2 = i. So i^z = i. But i is the fourth root of unity, making z = 4n | n ∈ ℤ. Optionally (1+ i) = √2*exp(iπ/4) and (1 - i) = √2*exp(-iπ/4), So (1 + i)/(1 - i) = exp(iπ/4 + iπ/4) = exp(iπ/2) = i, making z = 4n | n ∈ ℤ. Not sure how that can last 8 minutes.

[@trojanleo123]

z = 4n where n ∈ ℤ