Another Very Exponential Equation | Problem 364

[@orkunseyidoglu9089]

hocam çok teşekkür ederim sizin kanalı izleyerek karmaşık sayıları çok geliştirdim.

[@aplusbi]

Sevindim. Rica ederim

[@scottleung9587]

Nice - although I do have one question: why did you express 2^z as (e^ln(2))^z instead of (2*e^(2*pi*n*i))^z?

[@aplusbi]

Thanks!
That's a good question. I think it's because z is the exponent and that takes care of multiple values for the exponent to produce -sqrt(2). WA also agrees with this. If I include 2πni on the left hand side, it does not produce -sqrt(2). check this out:
www.wolframalpha.com/input?i2d=true&i=Power%5B2%2CDivide%5BlnSqrt%5B2%5D%2B%CF%80i%2Cln2%2B2%CF%80i%5D%5D
Funny that it rather gives us sqrt(2). I'm not exactly sure why this is happening (translation: too lazy to check 😜)

[@seanoliver2028]

I am still learning quadratics so i may be a bit off, but can't you just square both sides to get 2^(2z)=2 , and thus z=1/2 ?

[@seanoliver2028]

More specifically, could you not square the negative of a square root, -sqrt(2) , to get the square, 2 ?

[@XJWill1]

No, the log of the product equals the sum of the logs is NOT an identity for complex numbers. It is not even an identity for negative real numbers.

[@aplusbi]

counter-examples?

[@XJWill1]

@@aplusbi I'm sure you can find some yourself. You have a complex number YT channel after all!

[@GustavoHenrique-xp5wm]

​@@XJWill1Even if you're right, you sound like the idiot here

[@SweetSorrow777]

Rewrite the right-hand side as -1*(2^(1/2)) and -1 as e^(PI *i). The left-hand side as e^(z*ln2).
EZPZ

[@aplusbi]

lemon squeezy 😜

[@FisicTrapella]

So, the imaginary part of z gives us the minus sign...
We have 2^[(2n+1)(pi)i / ln2] = -1 similarly as e^[(2n+1)(pi)i ] = e^[(2n+1)(pi)i / lne] = -1

[@aplusbi]

The minus sign is a key to the imaginary world...😜😁