\[z_1 = a+bi, z_2 = c+di, \bar{z_1}=a-bi, \bar{z_2}=c-di\]
\[\bar{z_1}\bar{z_2}=(a-bi)(c-di)=(ac-bd)-i(bc+ad)=\overline{z_1z_2} \]
\[1/\bar{z}=1/(a-bi)=(a+bi)/(a^2+b^2)\]
\[1/z=1/(a+bi)=(a-bi)/(a^2+b^2), \overline{1/z}=(a+bi)/(a^2+b^2)\] #e
\[(1+i)^{4n}=(\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))^{4n}=2^{2n}(\cos(\pi n)+i\sin(\pi n))=2^{2n}(-1)^n\] \[(1+i)^{4n}=(\sqrt{2}e^{i\pi/4})^{4n}=2^{2n}e^{i\pi n}=2^{2n}(-1)^n\] # 4.9
\[|z_1 \pm z_2| \le |z_1|+|z_2|\]
\[z_1=a+ib, z_2=c+id\] Using Minkovsky inequality for sums:
\[|z_1+z_2|=\sqrt{(a+c)^2+(b+d)^2}\]
Using Minkovsky inequality for sums:
\[\left( \sum_{k=1}^n(a_k+b_k)^p\right)^{1/p}\le\left(\sum_{k=1}^n a_k^p\right)^{1/p} + \left(\sum_{k=1}^n b_k^p\right)^{1/p}\]
\[a_1=a, b_1=c, a_2=b, b_2 = d\] \[|z_1+z_2|=\sqrt{(a+c)^2+(b+d)^2}\le\sqrt{a^2+b^2}+\sqrt{c^2+d^2}=|z_1|+|z_2|\] \[a_1=a, b_1=-c, a_2=b, b_2 = -d\] \[|z_1+z_2|=\sqrt{(a-c)^2+(b-d)^2}\le\sqrt{a^2+b^2}+\sqrt{(-c)^2+(-d)^2}=|z_1|+|z_2|\] ##b) \[\left||z_1| -|z_2|\right| \le |z_1\pm z_2|\]
first prove: \[|z_1 + z_2|\ge |z_1|-|z_2|\]
\[z_3 := z_1+ z_2\] \[|z_3|+|-z_2|\ge|z_3+(-z_2)|\] \[|z_3|+|z_2|\ge|z_3-z_2|\] \[|z_1+z_2|+|z_2|\ge|z_1|\] \[|z_1+z_2|\ge|z_1|-|z_2|\] if \(z_4:=-z2\) then
\[|z_1-z_4|\ge|z_1|-|z_4|\] \[|z_1-z_2|\ge|z_1|-|z_2|\] so
\[|z_1\pm z_2|\ge|z_1|-|z_2|\] \[|z_1\pm z_2|\ge|z_2|-|z_1|=-(|z_1|-|z_2|)\] since \[|x|\le y \iff \begin{equation} \begin{cases} x \le y\\ -x \le y \end{cases}\,. \end{equation}\]
then \[|z_1\pm z_2|\ge||z_1|-|z_2||\] finally
\[||z_1|-|z_2||\le |z_1\pm z_2|\le|z_1|+|z_2|\] when \(z_1=0 \land z_2=0\)
\[||z_1|-|z_2||\le |z_1\pm z_2|\]
Find the locus of points lying on the complex plane complying to the following condition: \[\frac{|z-z_1|}{|z-z_2|}=n,\space z_1, z_2\in \mathbb{C}, n\in \mathbb{N}\]
Given the condition:
\[ \frac{|z - z_1|}{|z - z_2|} = n, \]
where \(z_1, z_2 \in \mathbb{C}\) and \(n \in \mathbb{N}\), we need to find the locus of points \(z\) on the complex plane.
The given equation can be rewritten as:
\[ |z - z_1| = n |z - z_2|. \]
This tells us that the ratio of the distances from \(z\) to \(z_1\) and \(z\) to \(z_2\) is \(n\).
This condition represents a locus known as an Apollonius circle. In general, for two fixed points \(A\) and \(B\) and a constant \(k\), the set of points \(P\) such that the ratio of distances \(\frac{|PA|}{|PB|} = k\) describes a circle (or a line if \(k = 1\)).
Let \(z_1 = x_1 + y_1i\) and \(z_2 = x_2 + y_2i\). The coordinates for \(z\) will be \(x + yi\).
Using the distance formula in the complex plane, we rewrite the condition as:
\[ |z - z_1| = n |z - z_2|. \]
Expanding this:
\[ \sqrt{(x - x_1)^2 + (y - y_1)^2} = n \sqrt{(x - x_2)^2 + (y - y_2)^2}. \]
Square both sides to eliminate the square roots:
\[ (x - x_1)^2 + (y - y_1)^2 = n^2 \left( (x - x_2)^2 + (y - y_2)^2 \right). \]
Expanding both sides:
\[ (x - x_1)^2 + (y - y_1)^2 = n^2 ( (x - x_2)^2 + (y - y_2)^2 ). \]
Expand and simplify each side:
\[ (x^2 - 2x x_1 + x_1^2) + (y^2 - 2y y_1 + y_1^2) = n^2 (x^2 - 2x x_2 + x_2^2) + n^2 (y^2 - 2y y_2 + y_2^2). \]
Combine like terms and rearrange the equation:
\[ x^2 + y^2 - 2x x_1 - 2y y_1 + x_1^2 + y_1^2 = n^2 x^2 + n^2 y^2 - 2 n^2 x x_2 - 2 n^2 y y_2 + n^2 x_2^2 + n^2 y_2^2. \]
Reorder the terms to get all terms involving \(x\) and \(y\) on one side:
\[ x^2 + y^2 - n^2 x^2 - n^2 y^2 - 2 x x_1 - 2 y y_1 + 2 n^2 x x_2 + 2 n^2 y y_2 = n^2 x_2^2 + n^2 y_2^2 - x_1^2 - y_1^2. \]
Factorize the terms involving \(x^2\) and \(y^2\):
\[ (1 - n^2) (x^2 + y^2) + 2 n^2 x x_2 + 2 n^2 y y_2 - 2 x x_1 - 2 y y_1 = n^2 (x_2^2 + y_2^2) - (x_1^2 + y_1^2). \]
To make this equation look like the standard form of a circle or determine its nature, solve for \(x\) and \(y\):
\[ (1 - n^2) x^2 + (1 - n^2) y^2 + 2 n^2 x x_2 + 2 n^2 y y_2 - 2 x x_1 - 2 y y_1 = n^2 (x_2^2 + y_2^2) - (x_1^2 + y_1^2). \]
Completing the square for the \(x\) and \(y\) terms will give the equation the standard form of a circle. However, note that if \(n = 1\), the equation simplifies significantly and describes a perpendicular bisector (a straight line).
Depending on the value of \(n\) and the positions \(z_1\) and \(z_2\), the set of points \(z\) satisfying the given condition represents an Apollonius circle in the complex plane. The explicit form of this circle can be calculated from the general derivation, confirming that the circle’s center and radius are derived based on the parameters \(n\), \(z_1\), and \(z_2\).
z2=(z)^3 z^4=((z)2)^3
z*2=(z)3 z4=z9
z^4=0
z^5=1 zero and unit roots of 5th degree
z^
z*= z^2/3 z1/3=z3
z^2=(z*)^3
\[z^n=\bar z\] \[\rho ^n e^{in\alpha}=e^{-i\alpha}\] ## Option 1. \(rho=0 \implies z = 0\)
\[e^{i(n-1)\alpha}=1\] \[\alpha = 2\pi k/(n-1), \space k \in \mathbb{N}\cup\{0\}, n\in \mathbb{N}\] or \[\alpha = 2\pi (k - 1)/(n-1), \space k, n\in \mathbb{N}\] #4.12 pe
\[|z|^2+3z+3\bar z=0\] \[\rho^2+3\rho e^{ia}+3\rho e^{-ia}=0\] \[\rho = 0\] \[\rho + 3e^{i\alpha}+3e^{-i\alpha}=0\] \[\rho + 6cos\alpha=0\] circle
eimz=1 z=2pk/m
1 * 2pi * 1 / m * 2 * pi * 2
product = (2pi)^(m-1)*(m-1)!/m
To verify whether the equality \((z^m)^{1/nm} = z^{1/n}\) holds for \(z \in \mathbb{C}\) and \(m, n \in \mathbb{N}\), we need to carefully interpret the expressions involving roots of complex numbers.
First, let’s rewrite the expressions using properties of exponents.
\[ (z^m)^{1/nm} = z^{m \cdot (1/nm)} = z^{1/n}. \]
This simplification follows directly from the rules of exponents:
\[ (z^k)^p = z^{kp}. \]
While the above algebraic manipulation seems straightforward, complex roots introduce multi-valued functions. Let’s delve deeper into the multi-valued nature of roots in the complex plane.
By definition, for \(z \neq 0\), a complex number \(z\) can be expressed in polar form as \(z = re^{i\theta}\), where \(r \geq 0\) (the modulus) and \(\theta\) (the argument) is any real number. Applying this to the expressions we have:
On the surface, these look the same: \(r^{1/n} e^{i \theta/n}\). However, due to the multi-valued nature of the argument, the argument \(\theta\) can be interpreted as \(\theta + 2k\pi\) for any integer \(k\). Hence:
For \((z^m)^{1/nm}\): Consider \(\theta_m = m \theta + 2k\pi\), then \((z^m)^{1/nm} = \left( re^{i(m \theta + 2k\pi)} \right)^{1/nm} = r^{1/n} e^{i(\theta + 2k\pi/m)/n} = r^{1/n} e^{i \theta/n + 2k\pi/nm}\).
For \(z^{1/n}\): The possible arguments for \(z\) are \(re^{i(\theta + 2l\pi)}\) for any integer \(l\). We get: \[ z^{1/n} = \left( re^{i(\theta + 2l\pi)} \right)^{1/n} = r^{1/n} e^{i(\theta + 2l\pi)/n}. \]
From the above: - \((z^m)^{1/nm} = r^{1/n} e^{i \theta/n + 2k\pi/nm}\) could yield \(nm\) distinct values. - \(z^{1/n} = r^{1/n} e^{i(\theta + 2l\pi)/n}\) could yield \(n\) distinct values.
Although simplification by algebraic property seems correct, multi-values root considerations impact equality assessment. Cases where multi-values align: - Given \(l\), distinct values for powers in both results should reflect similar roots cyclically for integers ensuring equal holds.
Despite complex root multiplicity, equal due exponents:
\[ \boxed{(z^m)^{1/nm} = z^{1/n}} \]
Holds true complex \(z\), \(m,n\in \mathbb{N}\).