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4

Just use the fact that$$\frac1{a+i\sqrt5\,b}=\frac{a-i\sqrt5\,b}{\left(a+i\sqrt5\,b\right)\left(a-i\sqrt5\,b\right)}=\frac{a-i\sqrt5\,b}{a^2+5b^2}.$$

3

If $r$ is a small positive real, one branch of exponentiation gives$$(-r)^{-r}=(re^{i\pi})^{-r}=\underbrace{r^{-r}}_{\approx1}\underbrace{e^{-ir\pi}}_{\approx1-ir\pi}\approx1-ir\pi.$$

3

Note that $$z=\sqrt{\frac{3 \pm \sqrt{5}}{2}}= \pm \left(\frac{1\pm \sqrt{5}}{2} \right).$$ So $$z_1=2\frac{1+\sqrt{5}}{4} =2 \cos 36^0$$ Hence the other three roots.

3

You can get a faster result by using a different way to complete the square: split the middle term $$z^4-3z^2+1=(z^2-1)^2-z^2=(z^2-z-1)(z^2+z-1)$$ Now the remaining two quadratic equations are easy to solve. $$0=z^2-z-1\implies z=\frac12(1\pm\sqrt{5})\\ 0=z^2+z-1\implies z=\frac12(-1\pm\sqrt{5})\\$$ The fifth unit roots satisfy the equation $z^5-1=0$. ...

3

If you actually expand $(1+z+z^2)(1+z+z^3)(1+z+z^4) - z(1+z)$ out, you will get: $$z^9+z^8+2z^7+3z^6+4z^5+4z^4+4z^3+3z^2+2z+1$$ $$=5z^4+5z^3+5z^2+5z+5 \quad \text{(why?)}$$ $$=5 \left( \frac{z^5-1}{z-1} \right)$$ $$=0$$ Therefore $(1+z+z^2)(1+z+z^3)(1+z+z^4) = z(1+z)$. In addition, you can use your idea of making the substitution $t=1+z$, so that way you ...

3

Note that if $z=\cos(2\pi/5)+i\sin(2\pi/5)=e^{2\pi i/5}\not=1$ then $z^5=1$. Therefore, before taking the product, we write each factor in a convenient way: \begin{align} (1+z+z^2)&=\frac{1-z^3}{1-z}=\frac{z^2-1}{z^2(1-z)}=-\frac{1+z}{z^2},\\ (1+z+z^3)&=\frac{z^4(1+z+z^3)}{z^4}=\frac{1+z^2+z^4}{z^4}=\frac{1-z^6}{z^4(1-z^2)}=\frac{z(1-z)}{(1-z^2)}=\... 3 The equality that you have used is not correct.x^{4}-4x^{3}+2x^{2}+4 \ne (x-1)^{4}-4(x-1)^{2} + 7$$3 You need to be very careful with \sqrt{a} or a^{\frac{1}{2}} even with the real numbers and even more so with the complex numbers. With the real numbers, positive numbers have two square roots but it is easy to pick one as special (the positive one) and define \sqrt{a} or a^{\frac{1}{2}} to mean the positive square root. With some care, you can ... 3 If x=\exp\left(\frac{\pi i}3\right)+\exp\left(-\frac{\pi i}3\right), then$$x^3=-1-1+3x=3x-2.$$But the equation x^3=3x-2 only has 2 solutions: 1 and -2. And since it is clear that both \exp\left(\frac{\pi i}3\right) and \exp\left(-\frac{\pi i}3\right) have absolute value equal to 1 but one of them is 1, it is clear that their sum is not 2. ... 3 No. Consider the companion matrix of the polynomial p(x)=x^4-2x^3-2x+1. Its eigenvalues are the roots of p(x), two of which are complex (non-real) numbers with absolute value 1, none of which is a root of unity. 3 Evaluate first,$$\frac{\sin\frac{11\pi}{9}}{1+\cos\frac{11\pi}{9}} =\frac{2\sin\frac{11\pi}{18}\cos\frac{11\pi}{18}}{2\cos^2\frac{11\pi}{18}} =\frac{\sin\frac{11\pi}{18}}{\cos\frac{11\pi}{18}}=\tan\frac{11\pi}{18}$$So, since \sin\frac{11\pi}{9} is negative, i.e. the 4th quadrant, the principal argument is \frac{11\pi}{18}-\pi = -\frac{7\pi}{18}  Or, ... 2 Consider z^5-1=0 So the roots of$$0=\dfrac{z^5-1}{z-1}=z^4+z^3+z^2+z+1$$are e^{2i\pi r/5},r=1,2,3,4 As z\ne0, like Quadratic substitution question: applying substitution p=x+\frac1x to 2x^4+x^3-6x^2+x+2=0 , divide both sides by z^2 Replace z+\dfrac1z=w\implies w^2=? to find$$w^2-2+w+1=0$$whose roots are$$2\cos\dfrac{2r\pi}5;r=1,2$$... 2 Set y = z^2, your equation is then equivalent to$$y^2 -3y +1=0.$$This is a quadratic equation and can be solve with the quadratic formula that you mentioned:$$y_{1,2} = \dfrac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1}}{2}=\dfrac{3 \pm \sqrt{5}}{2}.$$Now observe that$$y_1 = \dfrac{3+\sqrt{5}}{2}= \dfrac{\frac{1}{2}+\sqrt{5} + \frac{5}{2}}{2} = \dfrac{\frac{1+2\...

2

You are not wrong. The only complex numbers which are equal to their conjugates are the real numbers. And $\frac1z\notin\mathbb R$.

2

I like your idea that first, you denote $$w=z^2-z.$$ Then, recall that in polar form, you have $$c = r e^{i \theta} ,$$ where $c \in \mathbb{C}$ is any complex number, $r \in \mathbb{R}$ is its distance from the origin, and $\theta \in [0,2\pi)$ is its phase. So now, you can first find the first root, i.e. $w_1$, in a "regular way": $$w_1 = \sqrt{81} = ... 2 By scaling (replacing z,z_j by \frac{z}{R_2}, \frac{z_j}{R_2}) it seems that you want to prove the inequality: z\ne w, |z|=r, |w|=\rho <1 implies \frac{|1-z\bar w|}{|z-w|}\ge \frac{1+r\rho}{r+\rho}. If we let z=re^{i\alpha}, w=\rho e^{i\beta} and we square, this translates to: \frac{1+(r\rho)^2-2r \rho cos(\alpha-\beta)}{r^2+\rho^2-2r \rho ... 2 If a polynomial has two conjugate roots, let x\pm iy, it must be a multiple of$$(z-x-iy)(z-x+iy)=z^2-2xz+x^2+y^2.Notice that all coefficients are real. In fact, the roots of a polynomial of real coefficients are either reals or pairs of complex conjugates. Such a polynomial can always be factored as the product of quadratic and linear factors with ... 2 Yes, your formula holds true. Speaking the "algebraic" language, we identify a set W of words over an alphabet A with the corresponding element \sum_{w\in W}w of a free algebra on A (in our case, the \mathbb{Z}[\omega]-algebra). If W_n corresponds to the set of "allowed" words over \{V,U,U^\dagger\} of length n (and we consider the empty ... 2 COMMENT.-These are the two real roots given by Wolfram. It is impossible that you can calculate them by simple means. The two non-real roots are equally complicated. 2 I think your 3rd statement is wrong. See, in complex numbers we don't multiply two surds as we do normally with a real numbers. Like, for example \sqrt{-1}×\sqrt{-1}=-1 not 1. We don't use the identity a^n×b^n=(ab)^n. We instead multiply conserving the minus sign or you can say conserving the iota like, \sqrt{-1}×\sqrt{-1}=\sqrt{(-1)^2}=-1. So in ... 2 Note that \begin{align}(e^{i\pi/3}+e^{-i\pi/3})^3&=e^{i\pi}+3e^{i\pi/3}+3e^{-i\pi/3}+e^{-i\pi}\\&=-1+3(e^{i\pi/3}+e^{-i\pi/3})-1\end{align}\implies(e^{i\pi/3}+e^{-i\pi/3})^3-3(e^{i\pi/3}+e^{-i\pi/3})+2=0.$$This cubic x^3-3x+2=(x+2)(x-1)^2 has two roots:$$e^{i\pi/3}+e^{-i\pi/3}=-2,1.$$The former cannot be attained as \pi/3 is not a multiple ... 2 A strengthened version of the Gershgorin circle theorem does guarantee that if the circles are grouped into disjoint clusters, the number of eigenvalues in each cluster are equal to the number of disks there. The proof is as follows: let A=(a_{ij}), D=diag(a_{ii}) and A=D+E. Define A_t=D+tE and imagine increasing t from 0 to 1, keeping track of the ... 2 From$$\rho^2 e^{2i\theta}+ 3\rho e^{-i\theta}=0$$You should have concluded$$\rho = 0 \quad \text{or} \quad \rho = -3e^{-3i\theta}$$2 It's much easier to rearrange to |z|\le|z-1|. Geometrically, the distance of z from 0 is less than or equal to the distance of z from 1 in the complex plane, so the complex numbers z satisfying this are those with \Re(z)\le\frac{1}{2}, or x\leq\frac{1}{2}. Alternatively, you can write z=x+iy in |z|\le|z-1| and squaring both sides gives ... 2 The condition \lvert z-u\rvert<\lvert z\rvert simply means that z is closer to u than to 0. So, consider the perpendicular bisector of the line segment joining u to 0. It divides \mathbb C into two half-planes. Now, take the half-plane that contains u. 2 Notice that |a|=|-a| and that |b|+|a| \geq |b+a| so$$|z_2+1|+|z_1z_2+1|\geq|(z_2+1)-(z_1z_2+1)|$$And for the second$$|z_2-z_1z_2| = |z_2(1-z_1)|= |z_2||1-z_1| = 1\cdot |1-z_1|$$2 Let h=|z|-1>0. Then, for n\ge 2,$$\left|\frac n{z^n}\right|=\frac n{(1+h)^n}<\frac n{1+nh+\binom n2 h^2}\to 0$$2 Let  O  be the origin,  C  be 1,  P = 1 + \cos\left(\frac{11\pi}{9}\right) + i \sin \left(\frac{11\pi}{9} \right) , and  R  be 2, all in the complex plane. Then, if you draw a picture of this, you see that  O, P, R  are all points on the circle of radius 1 centered at  C .  \angle RCP  is  -\frac{7\pi}{9} . By a basic result in geometry on ... 2 The principal value of Arg lies in (-\pi,\pi] and e^{2ni\pi}=1, e^{i\pi}=-1. Given that$$Z=2 \cos 110^0 ~e^{11i \pi/18}= |2\cos 110^0|~ (1) e^{11i\pi/18} \implies Arg Z= \pi+11\pi/18$$in order to bring this value in (-\pi, \pi] we add 2n\pi to it where n=\pm 1, \pm 2, \pm 3,..., Finally, the principal value of Arg Z is$$\pi+ \frac{11 \pi}{18}...

1

Hint You have $$\frac{1}{2} +i\frac{\sqrt{3}}{2} = e^{i \frac{\pi}{3}}$$ and therefore...

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