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5 votes

Explanation for why $(-1)^{-i} = e^\pi$?

By arbitrary convention, $~-1~$ is often set to $~e^{i\pi},~$ rather than $~e^{i(\pi + 2k\pi)} ~: ~k \in \Bbb{Z}.~$ Then $$(-1)^{-i} = \left[ ~e^{i\pi} ~\right]^{-i} = e^{-(i^2)\pi} = e^\pi.$$ The ...
user2661923's user avatar
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3 votes

What are the solutions of $z^2=-1/\overline{z}$

This equation is equivalent to $|z|^2z = -1$. So $z$ is real and negative. This means that the modulus, or in this case absolute value, satisfies $|z|=-z$. So this gives $z^3 = -1$ This has only one ...
jMdA's user avatar
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2 votes

Showing $\int_{-1}^{1}\ln \left( \frac{x+1}{x-1} \right) \left( x - \sqrt{x^2 - 1} \right) \, dx=\frac{\pi^2 + 4}{2}$

Let’s consider the hyperbolic function substitution $x=\cosh(i\theta)$. Then our indefinite integral is transformed into $$ \begin{aligned} I & = \int \ln \left(\frac{x+1}{x-1}\right)\left(x-\...
Lai's user avatar
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2 votes

How to find principal value of the cubic root?

According to a comment on the similar question, the first non-real root as you traverse counter clockwise from the positive real axis is chosen by Mathematica.
Zack Fisher's user avatar
1 vote

Using the residue theorem to compute two integrals

For the second integral write $\zeta:=z+1$. We then want to find the coefficient of $\frac{1}{\zeta}$ in $$ \begin{eqnarray} \sin\frac{\zeta-1}{\zeta} &=&\sin(1-\frac1\zeta)\\ &=&\sin ...
ancient mathematician's user avatar
1 vote

Why do cubic equations always have at least one real root, and why was it needed to introduce complex numbers?

You already had good answers, so here is just a hint to grasp the idea: I don't see why the reason is "since $y^3−py−q$ is positive for sufficiently large positive $y$ and negative for ...
Basj's user avatar
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1 vote

Trying to solve a complex number question

We have: $p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ $q=\frac{a}{b}+\frac{a}{c}+\frac{c}{a}, abc=1$ Consider, $r=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ We notice $qr=3+\frac{1}{a^3}+\...
Dhaval Bothra's user avatar
1 vote

Showing $\int_{-1}^{1}\ln \left( \frac{x+1}{x-1} \right) \left( x - \sqrt{x^2 - 1} \right) \, dx=\frac{\pi^2 + 4}{2}$

Mathematica splits the integrand into real and imaginary parts as $$\left(\pi\sqrt{1-x^2} - x \log \frac{1-x}{1+x}\right) + i \left(\pi x + \sqrt{1-x^2} \log \frac{1-x}{1+x}\right)$$ The imaginary ...
user170231's user avatar
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1 vote

Explanation for why $(-1)^{-i} = e^\pi$?

See my answer to this question. Complex exponential expressions are generally multivalued. This comes from the periodicity of the exponential function. For nonzero $z$, one defines $z^w$ as $e^{w\log ...
MPW's user avatar
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1 vote

For a matrix $A(z)$ that represents the operation of multiplication with a complex number $z$, what does it mean for $e^{A(z)t} = A(e^{zt})$?

Let’s write $z=re^{j\theta}$, then $$A(z) = \begin{bmatrix} r\cos(\theta) & -r\sin(\theta) \\ r\sin(\theta) & r\cos(\theta) \end{bmatrix} $$ We can show by induction that $$A^k(z) = A(z^k)$$ ...
orchi_d's user avatar
  • 291
1 vote

Introduction to the Binary Tetrahedral group and the 24-cell

Q1: The fact that the generators squared or cubed give -1 (which has order 2) means that their order must be twice these, 4 or 6, to get +1, the unit element. Q2: The general concept in group theory ...
R. J. Mathar's user avatar
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