Can You Solve A Nice Exponential Equation? | Problem 235

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[@innovbrain]

I love your arithmetic solutions. This question was too easy. Only add Ln to both sides then you get 0 for ln(1), which is Z.lne=0, lne equals 1 already so it's Z that equals zero because 0*1=0

[@stevestarcke]

This channel is so wonderful! It always makes my day to see what looks insoluble is actually easy. Here, I can visualize myself as Ramanujan: A brilliant mathematician with little formal education.

[@aplusbi]

Thank you! 😍

[@prateek1.9]

somebody in comments really said, pls solve : (a+bi)²=b+ai

[@Nobodyman181]

Pls solve (a+bi)²=b+ai in video pls🙏

[@Nobodyman181]

Very easy

[@aplusbi]

Thanks a lot 😊

[@seanfraser3125]

Here’s the way that I like to do it that doesn’t depend on log (since that would be kind of circular reasoning). Let z=x+iy where x and y are real. Then 1 = e^(x+iy) = e^x*e^(iy) = e^x*cosy + ie^x*siny Setting real and imaginary parts equal, we have the following equations: e^x*cosy = 1 e^x*siny = 0 Since e^x can never be 0, the second equation gives siny = 0, or y = n*pi, where n is an integer. Plugging this into the first equation we have e^x*(-1)^n = 1 Since e^x > 0 for all x, (-1)^n = 1, so n is an even integer. We thus have e^x = 1, so x = 0 (x must be real). Putting this together, writing n=2k since n is even, we have z = 2kpi*i, where k is any integer.

[@trojanleo123]

I believe you are describing the "1st method" that was used in this video.

[@seanfraser3125]

@@trojanleo123 I usually give my solutions before watching the video so that I’m unbiased. They usually overlap with a+bi’s.

[@hafez591]

z=2.pi.n.exp(i.(pi/2+2.pi.k))

[@leoheriban5809]

If an exp. equals 1 the exponent has to be 0

[@aplusbi]

that's it?

[@whyyat3470]

Umm... z = 0

[@aplusbi]

that's it?

[@trojanleo123]

z = 2πni where n ∈ ℤ