Complex numbers

$(2+i)(3-i)+(2+3 i)(3+4 i)$;

$(2+i)(3+7 i)-(1+2 i)(5+3 i)$;

$(4+i)(5+3 i)-(3+i)(3-i)$;

$\frac{(5+i)(7-6 i)}{3+i}$;

$\frac{(5+i)(3+5 i)}{2 i}$;

$\frac{(1+3 i)(8-i)}{(2+i)^{2}}$;

$\frac{(2+i)(4+i)}{1+i}$;

$\frac{(3-i)(1-4 i)}{2-i}$;

$(2+i)^{3}+(2-i)^{3}$;

$(3+i)^{3}-(3-i)^{3}$;

$\frac{(1+i)^{5}}{(1-i)^{3}}$;

$\left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)^{3}$;

$\left(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)^{3}$.

$(1+i)^{8 n}=2^{4 n}, \quad n \in \mathbb {Z}$;

$(1+i)^{4 n}=(-1) 2^{2 n}, \quad n \in \mathbb {Z}$.

$\begin{cases} (1+i) z_{1}+(1-i) z_{2}=1+i, \\ (1-i) z_{1}+(1+i) z_{2}=1+3 i ; \end{cases}$

$\begin{cases} i z_{1}+(1+i) z_{2}=2+2 i, \\ 2 i z_{1}+(3+2 i) z_{2}=5+3 i ; \end{cases}$

$\begin{cases} (1-i) z_{1}-3 i z_{2}=-i, \\ 2 z_{1}-(3+3 i) z_{2}=3-i ; \end{cases}$

$\begin{cases} 2 z_{1}-(2+i) z_{2}=-i, \\ (4-2 i) z_{1}-5 z_{2}=-1-2 i ; \end{cases}$

$\begin{cases} x+i y-2 z=10, \\ x-y+2 i z=20, \\ i x+3 i y-(1+i) z=30. \end{cases}$

$(2+i) x+(1+2 i) y=1-4 i$;

$(3+2 i) x+(1+3 i) y=4-9 i$.

a complex number $z$ is real if and only if $\bar{z}=z$;

a complex number $z$ is purely imaginary if and only if $\bar{z}=-z$.

the product of two complex numbers is real if and only if one of them differs from the conjugate of the other by a real factor;

the sum and the product of two complex numbers are both real if and only if the given numbers are either conjugate or both real.

to their square;

to their cube.

$z^{2}=i$;

$z^{2}=3-4 i$;

$z^{2}=5-12 i$;

$z^{2}-(1+i) z+6+3 i=0$;

$z^{2}-5 z+4+10 i=0$;

$z^{2}+(2 i-7) z+13-i=0$.

5 ;

$i$;

-2 ;

$-3 i$;

$1+i$;

$1-i$;

$1+i \sqrt{3}$;

$-1+i \sqrt{3}$;

$-1-i \sqrt{3}$;

$1-i \sqrt{3}$;

$\sqrt{3}+i$;

$-\sqrt{3}+i$;

$-\sqrt{3}-i$;

$\sqrt{3}-i$;

$1+i \frac{\sqrt{3}}{3}$;

$2+\sqrt{3}+i$;

$1-(2+\sqrt{3}) i$;

$\cos \alpha -i \sin \alpha$;

$\sin \alpha +i \cos \alpha$;

$\frac{1+i \operatorname {tg} \alpha }{1-i \operatorname {tg} \alpha }$;

$1+\cos \varphi +i \sin \varphi , \quad \varphi \in [-\pi , \pi ] ;$

$\frac{\cos \varphi +i \sin \varphi }{\cos \psi +i \sin \psi }$.

$(1+i)^{1000}$;

$(1+i \sqrt{3})^{150}$;

$(\sqrt{3}+i)^{30}$;

$\left(1+\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{24}$;

$(2-\sqrt{3}+i)^{12}$;

$\left(\frac{1-i \sqrt{3}}{1+i}\right)^{12}$;

$\left(\frac{\sqrt{3}+i}{1-i}\right)^{30}$;

$\frac{(-1+i \sqrt{3})^{15}}{(1-i)^{20}}+\frac{(-1-i \sqrt{3})^{15}}{(1+i)^{20}}$.

$\left|z\right|+z=8+4 i$;

$\left|z\right|-z=8+12 i$.

$\left|z_{1} \pm z_{2}\right| \leqslant \left|z_{1}\right|+\left|z_{2}\right|$;

$\left|\left|z_{1}\right|-\left|z_{2}\right|\right| \leqslant \left|z_{1} \pm z_{2}\right|$;

$\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|$ if and only if the vectors $z_{1}$ and $z_{2}$ have the same direction;

$\left|z_{1}+z_{2}\right|=\left|\left|z_{1}\right|-\left|z_{2}\right|\right|$ if and only if the vectors $z_{1}$ and $z_{2}$ have opposite directions.

if $\left|z\right|<1$, then $\left|z^{2}-z+i\right|<3$;

if $\left|z\right| \leqslant 2$, then $1 \leqslant \left|z^{2}-5\right| \leqslant 9$;

if $\left|z\right|<1 / 2$, then $\left|(1+i) z^{3}+i z\right|<3 / 4$.

$(1+i)^{n}$;

$\left(\frac{1-i \sqrt{3}}{2}\right)^{n}$;

$\left(\frac{1-i \operatorname {tg} \alpha }{1+i \operatorname {tg} \alpha }\right)^{n}$;

$(1+\cos \varphi +i \sin \varphi )^{n}$.

$\sin 4 x$;

$\cos 4 x ;$

$\sin 5 x$;

$\cos 5 x$.

$\cos n x=\sum_{k=0}^{[n / 2]}(-1)^{k}\binom {n}{2 k} \cos^{n-2 k} x \cdot \sin^{2 k} x$;

$\sin n x=\sum_{k=0}^{[(n-1) / 2]}(-1)^{k}\binom {n}{2 k+1} \cos^{n-2 k-1} x \cdot \sin^{2 k+1} x$.

$\sin^{4} x$;

$\cos^{4} x$;

$\sin^{5} x$;

$\cos^{5} x$.

$\cos^{2 m} x=\frac{1}{2^{2 m-1}}\left[\sum_{k=0}^{m-1}\binom {2 m}{k} \cos (2 m-2 k) x+\frac{1}{2}\binom {2 m}{m}\right]$;

$\cos^{2 m+1} x=\frac{1}{2^{2 m}} \sum_{k=0}^{m}\binom {2 m+1}{k} \cos (2 m+1-2 k) x$;

$\sin^{2 m} x=$

$\sin^{2 m+1} x=\frac{(-1)}{2^{2 m}} \sum_{k=0}^{m}(-1)^{k}\binom {2 m+1}{k} \sin (2 m+1-2 k) x$.

$\sqrt[n]{z^{n} w}=z \sqrt[n]{w}$;

$\sqrt[n]{-z^{n} w}=-z \sqrt[n]{w}$;

$\sqrt[n]{z w}=u \sqrt[n]{w}$, where $u$ is one of the values of $\sqrt[n]{z}$.

$\sqrt[6]{i}$;

$\sqrt[10]{512(1-i \sqrt{3})}$;

$\sqrt[8]{2 \sqrt{2}(1-i)}$;

$\sqrt[3]{1}$;

$\sqrt[4]{1}$;

$\sqrt[6]{1}$;

$\sqrt[3]{i}$;

$\sqrt[4]{-4}$;

$\sqrt[6]{64}$;

$\sqrt[8]{16}$;

$\sqrt[6]{-27}$;

$\sqrt[4]{8 \sqrt{3} i-8}$;

$\sqrt[4]{-72(1-i \sqrt{3})}$;

$\sqrt[3]{1+i}$;

$\sqrt[3]{2-2 i}$;

$\sqrt[4]{-\frac{18}{1+i \sqrt{3}}}$;

$\sqrt[4]{\frac{7-2 i}{1+i \sqrt{2}}+\frac{4+14 i}{\sqrt{2}+2 i}-(8-2 i)}$;

$\sqrt[3]{\frac{1-5 i}{1+i}-5 \frac{1+2 i}{2-i}+2}$;

$\sqrt[4]{\frac{-2+2 \sqrt{3} i}{2+i \sqrt{5}}-5 \frac{\sqrt{3}+i}{2 \sqrt{5}+5 i}}$.

$\cos \frac{2 \pi }{5}$;

$\sin \frac{2 \pi }{5} ;$

$\cos \frac{4 \pi }{5}$;

$\sin \frac{4 \pi }{5}$.

$(z+1)^{n}+(z-1)^{n}=0$;

$(z+1)^{n}-(z-1)^{n}=0$;

$(z+i)^{n}+(z-i)^{n}=0$.

$\sqrt[n]{1}=\left\{ \varepsilon_{0}, \varepsilon_{1}, \ldots , \varepsilon_{n-1}\right\}$;

$\varepsilon_{k}=\varepsilon_{1}^{k}(0 \leqslant k<n)$;

$\varepsilon_{k} \varepsilon_{l}= \begin{cases} \varepsilon_{k+l}, & \text{ if } k+l<n, \\ \varepsilon_{k+l-n}, & \text{ if } k+l \geqslant n \end{cases} \quad (0 \leqslant k<n, 0 \leqslant l<n)$;

the set $\mathrm{U}_{n}$ of roots of degree $n$ of unity is a cyclic group of order $n$ under multiplication;

every cyclic group of order $n$ is isomorphic to the group $\mathbf{U}_{n}$.

if the numbers $r$ and $s$ are coprime and $\alpha^{r}=\alpha^{s}=1$, then $\alpha =1$;

if $d$ is the greatest common divisor of the numbers $r$ and $s$, then $\mathbf{U}_{r} \cap \mathbf{U}_{s}=\mathrm{U}_{d}$;

if the numbers $r$ and $s$ are coprime, then every root of unity of degree $r s$ is uniquely represented as the product of a root of degree $r$ and a root of degree $s$.

$\varepsilon$ is a primitive root of unity of degree $n$;

the order of $\varepsilon$ in the group $\mathbf{U}_{n}$ equals $n$;

$\varepsilon$ is a generating element of the group $\mathbf{U}_{n}$.

Let $z$ be a primitive root of degree $n$ of 1. Compute

Let $z$ be a primitive root of degree $2 n$ of 1. Compute

Let $z$ be a root of 1 with $z^{n} \pm z^{m} \pm 1=0$. Find $n$ and $m$.

the number of primitive roots of degree $n$ of unity equals $\varphi (n)$ (see 1.4);

if the numbers $m$ and $n$ are coprime, then $\varphi (m n)=\varphi (m) \varphi (n)$.

$\sigma (1)=1$;

if $n>1$, then $\sum_{d \mid n} \sigma (d)=0$;

$\sigma (p)=-1$, if $p$ is a prime number;

$\sigma \left(p^{k}\right)=0$, if $p$ is a prime number, $k>1$;

$\sigma (r s)=\sigma (r) \cdot \sigma (s)$, if the numbers $r$ and $s$ are coprime;

the function $\sigma (n)$ coincides with the Möbius function $\mu (n)$.

1 ;

2 ;

3 ;

4;

6 ;

12 ;

$p$, where $p$ is a prime number;

$p^{k}$, where $p$ is a prime number, $k>1$.

$\prod_{d \mid n} \Phi_{d}(x)=x^{n}-1$;

$\Phi_{2 n}(x)=\Phi_{n}(-x) \quad (n$ is an odd number greater than $1)$;

$\Phi_{n}(x)=\prod_{d \mid n}\left(x^{d}-1\right)^{\mu (n / d)}$;

if $k$ is divisible by every prime divisor of $n$, then

if $n$ is divisible by a prime number $p$ and is not divisible by $p^{2}$, then

$10$;

$14$;

$15$;

$30$;

$36$;

$100$;

$216$;

$288$;

$1000$.

all coefficients are integers;

the leading coefficient equals 1;

the constant term equals $-1$ when $n=1$ and equals $1$ when $n>1$.

$1-\binom {n}{2}+\binom {n}{4}-\binom {n}{6}+\ldots$;

$\binom {n}{1}-\binom {n}{3}+\binom {n}{5}-\binom {n}{7}+\ldots$;

$1+\binom {n}{4}+\binom {n}{8}+\ldots ;$

$\binom {n}{1}+\binom {n}{5}+\binom {n}{9}+\ldots$

$\cos x+\cos 2 x+\ldots +\cos n x=\frac{\sin \frac{n x}{2} \cos \frac{(n+1) x}{2}}{\sin \frac{x}{2}} \quad (x \neq 2 k \pi , k \in \mathbb {Z})$;

$\sin x+\sin 2 x+\ldots +\sin n x=\frac{\sin \frac{n x}{2} \sin \frac{(n+1) x}{2}}{\sin \frac{x}{2}} \quad (x \neq 2 k \pi , k \in \mathbb {Z})$;

$\cos \frac{\pi }{n}+\cos \frac{3 \pi }{n}+\cos \frac{5 \pi }{n}+\ldots +\cos \frac{(2 n-1) \pi }{n}=0 ;$

$\sin \frac{\pi }{n}+\sin \frac{3 \pi }{n}+\sin \frac{5 \pi }{n}+\ldots +\sin \frac{(2 n-1) \pi }{n}=0$;

$\frac{1}{n} \sum_{k=0}^{n-1}\left(x+\varepsilon_{k} y\right)^{n}=x^{n}+y^{n} \quad \left(\varepsilon_{0}, \varepsilon_{1}, \ldots , \varepsilon_{n-1}\right.$ are the roots of degree $n$ of unity);

$x^{2 n+1}-1=(x-1) \prod_{k=1}^{n}\left(x^{2}-2 x \cos \frac{\pi k}{2 n+1}+1\right) ;$

$x^{2 n}-1=\left(x^{2}-1\right) \prod_{k=1}^{n-1}\left(x^{2}-2 x \cos \frac{\pi k}{n}+1\right)$;

$\prod_{k=1}^{n-1} \sin \frac{\pi k}{2 n}=\frac{\sqrt{n}}{2^{n-1}}$;

$\prod_{k=1}^{n} \sin \frac{\pi k}{2 n+1}=\frac{\sqrt{2 n+1}}{2^{n}}$.

$1+\binom {n}{3}+\binom {n}{6}+\ldots =\frac{1}{3}\left(2^{n}+2 \cos \frac{\pi n}{3}\right)$;

$\binom {n}{1}+\binom {n}{4}+\binom {n}{7}+\ldots =\frac{1}{3}\left(2^{n}+2 \cos \frac{(n-2) \pi }{4}\right)$;

$\binom {n}{2}+\binom {n}{5}+\binom {n}{8}+\ldots =\frac{1}{3}\left(2^{n}+2 \cos \frac{(n-4) \pi }{3}\right)$;

$2 \cos m x=(2 \cos x)^{m}-\frac{m}{1}(2 \cos x)^{m-2}+\frac{m(m-3)}{1 \cdot 2}(2 \cos x)^{m-4}+\ldots$