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: $\int \frac{1}{x^{2}+1}dx = \frac{1}{2i} \int \frac{1}{x-i} - \frac{1}{x+i}dx$

#Algebra #ComplexNumbers #Logarithms #Integration #Euler'sFormula(ComplexNumbers)

Easy Math Editor

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Comments

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 - 6 years, 11 months ago

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 - 6 years, 11 months ago

\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 - 6 years, 11 months ago

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 - 6 years, 11 months ago

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 - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$