Linked Questions

0
votes
1answer
200 views

Splitting the exponent of a negative number? [duplicate]

Recently I have been chewing on a bit of a paradox in my mind, and I'm trying to figure out what I'm doing wrong. I am not in high school anymore, I have just graduated from college, and I haven't ...
19
votes
7answers
2k views

How does advancing through the math major work? [closed]

I am an undergrad math major that just completed Calculus 3 last semester. This semester I signed up for Discrete Mathematics, and will be taking Intro to Advanced/Abstract Math next. Of course-- I ...
13
votes
4answers
1k views

Confused about complex numbers

I am confused about something: \begin{eqnarray} (e^{2 i \pi})^{0.5} = (e^{2 i \pi \cdot 0.5})= e^{i \pi}=-1 \end{eqnarray} but \begin{eqnarray} e^{2 i \pi}=1~ and~ 1^{0.5}=1 \end{eqnarray} ...
5
votes
4answers
1k views

Euler's formula, is this true? [duplicate]

*I've changed this question as below. Let me have a function such as $ f(k) = \exp(j 2 \pi k ) $, where $k$ is real value. Using Euler's formula, we can write $f(k)$ as below, $$ f(k) = \exp(j 2 \...
3
votes
3answers
6k views

How can $i^i = e^{-\pi/2}$ !!

I was asked a homework question: find $i^i$. The solution provided was as follows: Let $A = i^i$. $\log A = i \log i$. Now, $\log i = \log e^{i\pi/2} = \frac{i\pi}{2}$. So, $\log A = -\frac{\pi}{2}...
16
votes
1answer
568 views

Proving that $e^\pi=e^{-\pi}$

I've been stuck with this for a while now. I have this chain of reasoning that would imply $e^{-\pi}=e^\pi$, obviously false, since $e^\pi$ and $e^{-\pi}$ are two real distinct numbers and so I must ...
5
votes
4answers
465 views

How to define $x^a$ for arbitrary real numbers $x$ and $a$

Questions like this, which asks to solve $$x^{\frac43} = \frac{16}{81}$$ confuse me. The solution for real $x$ is $\pm \frac8{27}$. The question would presumably be a bit different to solve $$x^{\...
6
votes
3answers
823 views

Are complex numbers subject to different rules of math? [duplicate]

From what I know, the rule to distribute exponents is like: $$(a b)^x = a^x b^x$$ Thus, if $a = \sqrt 2$ and $b = \sqrt 3$, $ab = \sqrt 6$. However, the imaginary unit $i = \sqrt{-1}$ has a ...
1
vote
2answers
2k views

Can the argument of a logarithm and its base be negative at the same time?

I'm struggling to understand why a $\log_x y$ is only true for $y > 0$. I know that if $y$ was $0$ , $x$ could only be $0$, but if it was $0$, then $y$ could be anything. So we'll leave that ...
2
votes
1answer
631 views

Exponent Laws: Fractional Exponents and Negative Bases

I have read through this thread. It mentions that there a several conventions for rational exponents.The first condition for the first convention for exponent laws to apply is that the base be greater ...
1
vote
1answer
497 views

Square root branch cut

Consider the following expression: \begin{equation} \phi(\delta)=i\,\sqrt{-3+i\,\delta}, \end{equation} where $\delta$ is infinitesimal. If we choose a branch cut along the negative real axes, it ...
4
votes
2answers
61 views

Why is $x^{p/q}$ ill-defined for $x<0$.

This is probably a duplicate but I can't find, if you do let me know and I will delete. Why is $x^{p/q}$ ill-defined for $x<0$. I can see that it is, $(-1)^{1/3} \neq (-1)^{2/6}$, but why? I ...
1
vote
1answer
56 views

How to plot the equation$x(x^2-1)^\frac{1}{3}$?

I would like to plot the function $x(x^2-1)^\frac{1}{3}$. When I do it in a statistical computing package, I dont get anything between $-1$ and $1$. However, when I plot this on google, I get a ...
2
votes
2answers
84 views

“Complex” Roots, when can we compute result wihout imaginay numbers? [closed]

The question is, how can we know if we can solve a n-index root without imaginary numbers when n-index is a non integer number, but a real number? First, basic theory: $$\sqrt[2.0]{8} = 2.8284.....$$...
3
votes
0answers
93 views

$e^{i\theta}$ confusion [duplicate]

I learned the following $$ e^{ik2\pi}=1 $$ and I was wondering whether or not $k$ has to be an integer. Thought 1: Since $e^{ik2\pi}=\cos(2k\pi)+i\sin(2k\pi)=1$, equating the real and imaginary ...

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