Linked Questions

50 votes
2 answers

What is wrong with this fake proof $e^i = 1$?

$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$ Obviously, one of my algebraic manipulations is not valid.
user58789's user avatar
  • 511
5 votes
4 answers

Exponent of a number is a square root?

Say you have a number $x^{\sqrt 2}$. Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my ...
Qwerp-Derp's user avatar
7 votes
1 answer

Simple Complex Number Problem: $1 = -1$ [duplicate]

Possible Duplicate: -1 is not 1, so where is the mistake? I'm trying to understand the exact point of failure in the following reasoning: \begin{equation*} 1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{...
Sami's user avatar
  • 797
2 votes
3 answers

Why $\infty×0=-1$ from multiplication of two slopes of two lines perpendicular to each other and how do we define infinity?

Here is given $A(x_1,y_1), B(x_1,y_2), C(x_2,y_3)$ and $D(x_3,y_3)$. I have recently read that, multiplication of two perpendicular lines is always $-1$. From the above graph, the slope of $AB, m_1 = ...
Anirban Niloy's user avatar
1 vote
3 answers

What is wrong with the reasoning in $(-1)^ \frac{2}{4} = \sqrt[4]{(-1)^2} = \sqrt[4]{1} = 1$ and $(-1)^ \frac{2}{4} = (-1)^ \frac{1}{2} = i$?

$$(-1)^ \frac{2}{4} = \sqrt[4]{(-1)^2} = \sqrt[4]{1} = 1$$ $$(-1)^ \frac{2}{4} = (-1)^ \frac{1}{2} = i$$ Came across an interesting Y11 question that made pose this one to my self. I can't for the ...
allquiet1984's user avatar
1 vote
1 answer

Possible fake proof of $1= -1$ [duplicate]

Possible Duplicate: -1 is not 1, so where is the mistake? Simple Complex Number Problem: 1 = -1 Well, I remembered this after having Algebra II a year ago, is it possible that this is a valid ...
Rivasa's user avatar
  • 1,575
2 votes
4 answers

Why does $x^{\frac{1}{a}} = \sqrt[a]{x}$?

I was attempting to answer this question, and came across the fact that $x$ to a non integer power, $x^\frac{1}{b}$ is equal to $\sqrt[b]{x}$. Why does this work? Similarly, does $x^{\frac{a}{b}} = a\...
Travis's user avatar
  • 3,406
0 votes
4 answers

What are multiplicative inverses?

Why does $i^{-4} = -1$? This arrises from the accepted answer of is $i^4$ equal to $-1$? In which the answerer states: The powers of $i$ are cyclic ... the pattern persists for negative ...
Travis's user avatar
  • 3,406
3 votes
1 answer

Why are the graph of $f(x) = (x^2)^\frac{1}{6}$ and the graph of $f(x) = x^\frac{1}{3}$ not the same?

Why is the graph of $f(x) = (x^2)^\frac{1}{6}$ not the same graph as $f(x) = x^\frac{1}{3}$? Shouldn't they be the same because when you apply the exponent rules to the first equation, you get the ...
av44's user avatar
  • 31
2 votes
1 answer

Are exponents commutative? If so, doesn't that imply $1=-1$, since $((-1)^2)^{1/2}=((-1)^{1/2})^2$? [duplicate]

I have been studying maths for a couple years now, but even though this is a very basic question, I am surprised that it never crossed my mind before. I have been taught in school, that $(a^m)^n$ = $(...
Srinidhi kabra's user avatar
0 votes
1 answer

Complex integration and differentiation for $\gamma:[a,b]\rightarrow \mathbb{C}$

Let $\gamma(t):[a,b]\rightarrow \mathbb{C}$ , $[a,b]\subset \mathbb{R}$, $\gamma(t)= \gamma_1(t) + i\gamma_2(t)$ $\int_{a}^{b} \gamma(t) dt = \int_{a}^{b} \gamma_1(t) dt + i\int_{a}^{b} \gamma_2(t) ...
Little Rookie's user avatar
-2 votes
1 answer

Euler's identity confusion: $e^{\pi} = e^{-\pi}$? [duplicate]

We know from Euler's identity that $$e^{\pi i} = -1.$$ So that means $$(e^{\pi i})^{-i} = e^{\pi} = (-1)^{-i} = ((-1)^{-1})^{i} = (-1)^{i}.$$ But also we have $$ (-1)^{i} =( e^{\pi i})^{i} = e^{-\pi}...
Ruby's user avatar
  • 311
-3 votes
2 answers

How is $\mathrm{Log}(i^{i})=-i^{-1}\mathrm{Log}(-i)$ [duplicate]

Question: Show that $$\color{Blue}{e^{-\pi/2}=i^{i}}$$ My answer: First establish that $\qquad\qquad\quad e^{-i\pi/2}=\cos(-\pi/2)+i \sin(-\pi/2)=-i$ then $\qquad\qquad\qquad\qquad \mathrm{Log}(e^{-i ...
Isaac's user avatar
  • 11
0 votes
1 answer

Can $\lim_{h\to 0} (1+h)^{[\frac{x}{h}]}$ be an equivalent definition of the exponential function?

Here, take $[x]$ to be the smallest integer function. I like this more than $\lim_{n\to \infty} (1+\frac{x}{n})^n$ because it seems to make the property $\exp(x+y)=\exp(x)\exp(y)$ obvious. This is ...
Ryder Rude's user avatar
  • 1,417
1 vote
1 answer

understanding of $2^{-i} \cdot 3^{-i}=\left(\frac16\right)^i$

I'm having trouble understanding how $2^{-i} \cdot 3^{-i}$ reduces to $\left(\frac16\right)^i$ ? Could anyone assist understanding the rules behind this reduction?
edd91's user avatar
  • 27