# Tag Info

### 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 ...
• 37.3k
Accepted

### 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 ...
• 1,568
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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 ...
• 1,561
1 vote
Accepted

• 44.2k
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 $$A^k(z) = A(z^k)$$ ...
• 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 ...
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