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Math is getting broken.

the is some problem in log identities. please take a look : \[f(x) = log \frac{x-i}{x+i} \] \[ g(x) = log \frac{x+i}{x-i} \] just by inspection we can conclude very easily that \[f(x) = - g(x)\] but computing f(0) and g(0), we get \[f(0) = log \frac{0-i}{0+i} = log(-1)\] \[g(0) = log \frac{0+i}{0-i} = log(-1)\] for log(-1) we can use euler's identity that: \[ e^{i\theta} = cos\theta + isin\theta \] \[for \theta = \pi\] \[e^{i\pi} = -1\] taking log \[ ln(-1) = i\pi \] hence we got then : \[ f(0) = g(0) = i\pi \] this is contradicting f(x) = - g(x). i tried finding some limitations for loga - logb = log(a/b) identity on internet, but there are not.

so please tell me if there is some wrong in my computation or else.

i like complex numbers, and i think they are more than just imaginary.

by the way i got this problem while trying to compute the integral by partial fractions: \[ \int \frac{1}{x^{2}+1}dx = \frac{1}{2i} \int \frac{1}{x-i} - \frac{1}{x+i}dx \]

Note by Soham Zemse
3 years, 2 months ago

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\sqrt { a*b } =\sqrt { a } *\sqrt { b } holds if and only if any one of a and b is non negative.... thus your second step where you cancel -i/i to get -1 is wrong because both numerator and denominator have negative real numbers hence the rule stated earlier cannot be used...... hence value of f(0)= log(1) and not log(-1). Abhinav Raichur · 3 years, 2 months ago

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@Abhinav Raichur i just multiplied and divided by i in (-i/i) to get (1/-1) = -1. anyways thanks for that rule, it was troubling me. Soham Zemse · 3 years, 2 months ago

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@Soham Zemse try this now! {its an interesting paradox like the one you shared above}.. here's the link:

{ https://brilliant.org/discussions/thread/beware-of-blind-algebra/} .......... i would also request you to reshare this as i have a poor following :p........... thanks!! Abhinav Raichur · 3 years, 2 months ago

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Log doesn't apply to negative number.

What is log 0 ? ( let base be 10)

Log 0 = x => 0 = 10^x

X has to be -infinity for this to hold true.

So 0 is limit. You can't go below 0. Technically speaking log 0 is also undefined. And so is log (-1)

So our general rule won't hold true.

Limitation : log applies only to positive real numbers Sanath Kumar B P · 3 years, 2 months ago

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@Sanath Kumar B P hi,

yes in real calculus, log of negative number is not defined, but this doesn't stops us from seeking further,

you can have a read on http://en.wikipedia.org/wiki/Complex_logarithm

in fact when a number is not a positive real number, its logarithm is a complex number. eg : \[z=r e^{i\theta}\] then if we could take natural log (base e) on both side, \[ ln z = ln r + i\theta \] where r is the modulus of z and theta is principal argument. Soham Zemse · 3 years, 2 months ago

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