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Cosine of a multiple of arctangent

The method is basically the same as the one for computing "more standard" expressions like $\cos(\arctan x)$ for instance (cf. here). Start by constructing a right triangle with catheti of ...
Abezhiko's user avatar
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1 vote

Why is ${(e^{(i×1)})}^π$ different from $e^{(i×π)}$

First, \begin{align*} \mathrm{e}^{\mathrm{i}} &= 0.54\,030\,230\,586\,813\,971\,740\,093{\dots} \\ &\qquad + 0.84\,147\,098\,480\,789\,650\,665\,250{\dots}\mathrm{i} \end{align*} The ellipses ...
Eric Towers's user avatar
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1 vote

how to find expression of $a\text{ mod } n$ using polynomial of $e^{\frac{2a\pi i}{n}}$?

from comments I learnt about the Discrete Fourier Transform (DFT) which give us $$g(n,a)=\sum_{k=0}^{n-1} f(n,k) \exp\left(\frac{2ak \pi i}{n} \right) \leftrightarrow f(n,k)=\frac{1}{n}\sum_{p=0}^{n-1}...
Faoler's user avatar
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1 vote
Accepted

Prove that $A$ is a bounded automorphism of the vector space $X$. Also prove that $\sigma(A) = \{ \lambda \in \mathbb{C} \mid |\lambda| = 1 \}$.

When $c=0$, $A$ is the identity operator, whose spectrum is just $\{1\}$, so $\sigma(A)=S^1$ doesn't hold. We assume $c\not=0$ in the next. If $\lambda\in\sigma(A)$, $|\lambda|\le\|A\|=1$, and we also ...
Just a user's user avatar
0 votes

Construct an explicit biholomorphism between two domains

$f(z)=e^z+z$ is the biholomorphism from $D_2$ to $D_1$. This comes entirely from attention...
Fresh's user avatar
  • 81
1 vote
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What are the polynoms in $R_{n−1}[X]$ which are real-valued over the set of the n-ths roots of unity?

There are a lot more. For example, when $n$ is even, $\beta=ax+b$ (instead of being just a real number) works because the degree of $\prod_{\omega\in\mathbb U_n\setminus\mathbb R}(x-\omega)$ is $n-2$, ...
Just a user's user avatar
1 vote

Probable typo in a book for the complex limit $\lim_{z \to 1} \frac{z}{1+ \bar z}$

Set $z=1+w.$ Then $w\to 0.$ Hence we may assume that $|w|\le 1.$ Next $$\left |{z\over 1+\overline{z}}-{1\over 2}\right |=\left |{1+w\over 2+\overline{w}}-{1\over 2}\right |\\ ={|w+2\overline{w}|\over ...
Ryszard Szwarc's user avatar
4 votes
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Probable typo in a book for the complex limit $\lim_{z \to 1} \frac{z}{1+ \bar z}$

You do not need to write $z$ polar form. We have $$\lim_{z \to 1} \frac{z}{1+\bar z} = \frac{\lim_{z \to 1}z}{\lim_{z \to 1}1+\bar z} = \frac 1 2 .$$ If your book claims that the limit is $1$, then ...
Paul Frost's user avatar
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4 votes
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Solve simultaneous equation with 3 variable using Vieta relations

From $x+y+z=3$, we deduce then $$\implies \cases{x^2 =y^2+(3-x-y)-1\\y^2 =z^2+(3-y-z)-1\\z^2 =x^2+(3-z-x)-1}\implies \cases{x^2+x =y^2-y+2\\y^2+y =z^2-z+2\\z^2+z =x^2-x+2}\tag{1}$$ Subtract the first ...
NN2's user avatar
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5 votes
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In complex number system, sin z and cos z are unbounded and periodic. But they are continuous also. How can that be possible?

Put simply, they are periodic in one direction and unbounded in the other direction. Specifically, their values repeat if you shift by $2\pi$ along the real axis, but increase unboundedly along the ...
JonathanZ's user avatar
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2 votes

Solve $z^2+|z^2| = 1+2i$

Hint: Let $z^2=u$ $$u=1-|u|+2i$$ Equating the imaginary parts, $u$ can be written as $a+2i$ where $a$ is real $$\implies a=1-\sqrt{a^2+4}$$ $$\implies a^2+4=(1-a)^2\iff2a=-3$$ $$\implies z^2=u=?, z=?$$...
lab bhattacharjee's user avatar
2 votes
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Why is this particular substitution made? ($y=tx$)

It is a common trick to solve equations that have terms of the form $x^ky^{n-k}$ (the sum of powers of $x$ and $y$ is the same across all the terms). Do the $y=0$ case separately, then divide by $y^n$ ...
ultralegend5385's user avatar
1 vote

Solving a System of Equations Involving Complex Variables and Their Magnitudes

You can start by eliminating $y = c_1 - x$. The remaining equation is $$ |x| + |c_1 - x| = r$$ Think of this geometrically in the complex plane. $|x| = s$ says $x$ is on the circle of radius $s$ ...
Robert Israel's user avatar
5 votes
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Example of a complex function $f: \mathbb{C} \rightarrow \mathbb C$ differentiable only at $z_1=1+i\,, z_2=1-i\,, z_3=-1+i\,, z_4= -1-i$

This is easy to do using the Wirtinger derivatives. Let$$\varphi(z)=(z-1-i)(z-1+i)(z+1-i)(z+1+i)=z^4+4.$$Then you are after a function $f$ such that $\frac\partial{\partial\overline z}f(z)=\varphi(z)$....
José Carlos Santos's user avatar
1 vote

Example of a complex function $f: \mathbb{C} \rightarrow \mathbb C$ differentiable only at $z_1=1+i\,, z_2=1-i\,, z_3=-1+i\,, z_4= -1-i$

These $4$ points in $\mathbb R^2$ are characterized by an equation such as $$(x^2-1)^2+(y^2-1)^2=0.$$ If we set $u(x,y)=f(x)$ and $v(x,y)=g(y)$ then we already have $u_y=-v_x=0$ everywhere. Now choose ...
Lieven's user avatar
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1 vote

why does $|z|=\sqrt{\Re^2(z)+\Im^2(z)}$, which elements to $\mathbb{R}$?

This is an unusual situation, where the original poster is asking for an intuitive explanation for the formula for $~|z| ~: ~z \in \Bbb{C}.~$ Given this situation, I don't see how that the OP (i.e. ...
user2661923's user avatar
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1 vote
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Relationship between the parts of a complex number with its conjugate

This requires some algebraic background, but yes. In general, if $L/K$ is a finite field extension, there are two special functions from $L$ to $K$ called the field trace $\text{tr}_{L/K}$ and the ...
Qiaochu Yuan's user avatar
4 votes
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How does this solution involving complex numbers work on this inequality?

Choose $\omega=e^{2\pi i/5}$, we verify the three equations. First equation: \begin{align*} \sum_{j=1}^5|z_j|^2&=\frac15\sum_{j=1}^5\left|\sum_{k=1}^5a_k\omega^{-jk}\right|^2\\ &=\frac15\sum_{...
Mengchun Zhang's user avatar
3 votes
Accepted

Area & Perimeter of region $S$ containing points of the form $a+b\omega+c\omega^2$ where $a,b,c \in [0 , 1], \omega=-\frac 12+i\frac{\sqrt 3}{2}$

I won't give you a complete answer, but I will give you an outline for reasoning in an efficient way. Suppose you have two vectors $\vec u$, $\vec v$ that are not parallel. Then for $b, c \in [0,1]$,...
heropup's user avatar
  • 140k
3 votes
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Let $a,b,c,d\in\mathbb{C}$ such that $|a|+|b|\leq 1$ and $|c|+|d|\leq 1$. Show that $|3a+b+3c-d|+|a+3b-c+3d|\leq 7$.

I think I have a full solution. Thanks to River Li for pointing me in the right direction. We can reparameterize as $$a=r_1e^{i\theta_1},\\b=r_2e^{i\theta_2},\\c=r_3e^{i\theta_3},\\d=r_4e^{i\theta_4}$$...
w.w's user avatar
  • 101
2 votes
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How to show that the abstract complexification of a real matrix Lie algebra (vector space) coincides with complex linear combinations?

I don't think this has anything to do with Lie algebras, this is just linear algebra. Let $K\subseteq L$ be a field extension, let $V$ be an $L$-vector space and $W$ be a $K$-subspace of $V$. We have ...
tomasz's user avatar
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1 vote

Let $a,b,c,d\in\mathbb{C}$ such that $|a|+|b|\leq 1$ and $|c|+|d|\leq 1$. Show that $|3a+b+3c-d|+|a+3b-c+3d|\leq 7$.

Just some thoughts. (i) I guess the maximum is $2\sqrt{10}$ attained at e.g. $$a = \mathrm{e}^{\mathrm{i}\theta},\quad b = c = 0, \quad d = \mathrm{i} \mathrm{e}^{\mathrm{i}\theta}, \quad \forall \...
River Li's user avatar
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0 votes

Spivak "Calculus" understanding authors intend.

Let $B = \{ \tau \in [0,2\pi) \mid f(\cos\theta+i\sin\theta) = \cos\frac{\theta}{2}+i\sin\frac{\theta}{2} \text{ for all } \theta \in [0,\tau] \}$. In particular, if $\tau \in B$, then $f(\cos\tau+i\...
Kritiker der Elche's user avatar
1 vote
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Are complex numbers considered as scalars or vectors?

Elements of sets can be both vectors and scalars depending on the context. Vectors are always elements of a vector space and scalars are always elements of the corresponding field. Some examples: The ...
spinosarus123's user avatar
0 votes

Need help in explaining weirdness when representing complex numbers using vectors

You found out that if we have two complex numbers $z_{1}= (a + bi) $ and $z_{2}= (c+di)$ that the product $ac - bd + ( bc +ad)i $ is not what you want for the dot product. You would like $ac + ...
kirk beatty's user avatar
0 votes

How to correctly find the angle phi in de moivre's formula?

$\phi$ must be an angle such that $a = r\cos\phi$ and $b = r\sin\phi.$ The value of the arc tangent function is strictly between $-\dfrac\pi2$ and $\dfrac\pi2$ for every possible input. That is, it ...
David K's user avatar
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0 votes

Solve $\cos{z}+\sin{z}=2$

For real $z$, there is no solution. For complex number $z$, we let $t=\tan \frac z2$, then the equation becomes a quadratic equation. $\displaystyle \begin{aligned}& \frac{1-t^2}{1+t^2}+\frac{2 t}{...
Lai's user avatar
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2 votes

Quicker way to get to $\sin^n(\theta)$?

For all $n \in \mathbb {N}$, $ \begin{align} {\sin}^{2 n} \left( \theta \right) & = {\left( \frac {{e}^{i \theta} - {e}^{- i \theta}}{2 i} \right)}^{2 n} \\ & = \frac {1}{{\left( 2 i \right)}^{...
Simon's user avatar
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3 votes

Need help in explaining weirdness when representing complex numbers using vectors

I prefer to discuss the issue differently. Given any complex number $~z \in \Bbb{C},~$ you can uniquely identify $~x,y \in \Bbb{R},~$ such that $~z = x + iy ~: ~i = \displaystyle \sqrt{-1}.$ Note ...
user2661923's user avatar
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10 votes
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Need help in explaining weirdness when representing complex numbers using vectors

The heart of your issue is this: the dot product of two complex numbers and the usual product of complex numbers is not the same product. Full stop. There is no "in some cases it's the same and ...
Vercassivelaunos's user avatar
2 votes

Need help in explaining weirdness when representing complex numbers using vectors

Your Inconsistency When you calculate the magnitude $$ \left|\vec{OM}\cdot\vec{OM}\right| = 3^{2} + 4^{2} $$ you multiply the real part of the vector with the real part, the imaginary part with the ...
acat3's user avatar
  • 12k
2 votes

Let $f(x)=a_0+a_1x+\cdots +a_{2m}x^{2m}$ be a polynomial with $a_0=a_{2m}=1$.

You're close with what you have. Observe that $ f(x) = (x^2 + ax + 1) ^m$, where $ a \in (-2, 2)$. Use the multinomial theorem to show that $ a_m = g(a, m), $ where $g$ is a polynomial in $a$. Hence $...
Calvin Lin's user avatar
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2 votes
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angle of $\frac{\sum_i a_i e^{j\theta_i}}{\sum_i a_i}$

Example: $a_1 = e^{i \phi}$ and $a_2 = -e^{-i\phi}$ with some “small” $\phi > 0$. Then $$\arg(a_1 + a_2) = \pi/2$$ and $$ \arg(e^{-2i \phi} a_1 + e^{2i \phi} a_2) = \arg(\overline{a_1} + \overline{...
Martin R's user avatar
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2 votes
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Prove this formula results in an integer for every $n$ in $\mathbb N$

Let $a_n=\left( \dfrac{36 + 45 \sqrt{2} i}{22} \left( {1 + 2 \sqrt{2} i} \right)^n + \dfrac{36 - 45 \sqrt{2} i}{22} \left( {1 - 2 \sqrt{2} i} \right)^n \right) \cdot (-1)^{n+1} \cdot \dfrac{1}{9} + \...
J. W. Tanner's user avatar
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1 vote

What are uses of complex numbers?

The complex exponential links the trigonometric functions to the complex plane. $e^{ix} = \cos x + i\sin x$ Because the exponential function is so easy to work with, the complex exponential appears in ...
user317176's user avatar
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6 votes

Find the maximum value of $|z|$ if $|z^2+1|=2|z+1|$

Put $z := x + iy$, and square both sides to get $$|z^2+1|^2 = 4|z+1|^2.$$ Simplifying, this becomes $$(x^2+y^2)^2 = 2x^2 + 6y^2 + 8x + 3. \tag{1}$$ Now, if we set $x^2 + y^2 = R^2$ (geometrically, ...
stoic-santiago's user avatar
1 vote
Accepted

Need help explaining why the area of a triangle remains constant regardless of the positions of two points

Modifying answer as ten_to_tenth saw that answer could be in terms of linear combination. I like that approach so here is a changed answer. (The central idea of rotating a triangle around one of its ...
kirk beatty's user avatar
1 vote

Need help explaining why the area of a triangle remains constant regardless of the positions of two points

If $A = w_1$ and $B = w_2$, then we may take $w_1 = 1 + i + \sqrt{2} e^{i \theta}$ and $w_2 = 1 + i + \sqrt{2} e^{i \varphi}$ where $\theta, \varphi \in [0, 2\pi)$ and $|\theta - \varphi| = \pi/2$. ...
heropup's user avatar
  • 140k
1 vote

Seeking alternative approaches: $w = \dfrac{12z}{|z^2|}$ with $|z| = |z - 6 - 6i|.$ Show that $w$ lies on a circle.

Alternative, simpler but less elegant approach, that may be viewed as a rehabilitation of the $~(x + iy)~$ approach tried by the original poster: First, setting $~z = x + iy ~: x,y \in \Bbb{R}, ~$ ...
user2661923's user avatar
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1 vote
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Seeking alternative approaches: $w = \dfrac{12z}{|z^2|}$ with $|z| = |z - 6 - 6i|.$ Show that $w$ lies on a circle.

The difficulty you're having seems to arise from not actually using the relationship between $x$ and $y$ that restricts $z$ to a line. If $z = x + yi$ lies on a line satisfying $x + y = 6$, then $z = ...
heropup's user avatar
  • 140k
-1 votes

Divergence of the generalized continued fraction $1+ \frac{-1\mid}{\mid1}+\frac{-1\mid}{\mid1}+\frac{-1\mid}{\mid1}+\dots$

$$a=1-\frac{1}{a}$$ $$a^{2}=a-1$$ $$a^{2}-a+1=0$$ Using quadratic equation: $$a=0.5\pm \frac{i\sqrt{3}}{2}$$ Hence I think this converges?
hablahblahha's user avatar
0 votes

Equation in space of complex numbers

First of all, the roots can be $3$ and $5$ (real but not equal) and so you need to consider $D\neq0$ not $D<0$. One way to proceed is: Since $z_1$ and $z_2$ are roots of the given quadratic, $z_1+2$...
ultralegend5385's user avatar
1 vote

If a root of the equation $az^2+bz+c = 0$ has modulus 1, then $b^2=ac$.

${\bf Solution \, 1.}$ Dividing through by $a$, you can assume $a = 1$ and $|b| = |c| = 1$; this does affect whether or not $b^2 = ac$. Writing $c = e^{i\alpha}$, you can also replace $z$ by $ze^{i{\...
Zarrax's user avatar
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4 votes

If a root of the equation $az^2+bz+c = 0$ has modulus 1, then $b^2=ac$.

Let $z$ denote the root of the equation and $|z|=1.$ We have $$ab^{-1}z+1+cb^{-1}z^{-1}=0$$ Hence the imaginary parts of the numbers $ab^{-1}z$ and $cb^{-1}z^{-1}$ are opposite. As both numbers have ...
Ryszard Szwarc's user avatar
0 votes

How does raising a complex number to n affect its principal argument

Yes, your posted analysis is accurate, and is consistent with De Moivre's formula. Specifically, $~\displaystyle \left[ ~re^{i\theta} ~\right]^5 = r^5 \times e^{i5\theta}.$ Your only mis-step seems to ...
user2661923's user avatar
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0 votes

How does raising a complex number to n affect its principal argument

If we are raising to the $n$-th power the argument changes by scaling: $$ \theta \to n\theta $$ But a rotation of $2\pi$ results in the same thing in the argand diagram, hence $z$ will lie in the $c$ ...
Masd's user avatar
  • 886
1 vote

Complex numbers using hence result

Write $\zeta:=\exp{\frac{2\pi i}{12}}$. The given identity is $\zeta(1+\zeta^{-1})=1+\zeta$. Now $$ \begin{align} &(1+\zeta)^6+(1+\zeta^{-1})^6\\ &=\zeta^6(1+\zeta^{-1})^6+(1+\zeta^{-1})^6\\ &...
ancient mathematician's user avatar

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