Вычислите выражения \((9+i)^{10}\) и \(\frac{9 + 2i}{9 + i}\)
\[
|z| = |9 + i| = \sqrt{1 + 81} = \sqrt{82}
\] \[
\arg{z} =arctg \frac{1}{9} = 0.1106572
\] \[
z = \sqrt{82} * (cos(0.1106572) + i*sin(0.1106572))
\]
\[
z^{10} = (\sqrt{82})^{10} * (cos(0.1106572 * 10) + i*sin(0.1106572 * 10)) = 1659911102 + 3315041217i
\]
\[\frac{9 + 2i}{9 + i} = \frac{(9+2i)(9-i)}{(9+i)(9-i)} = \frac{(9*9 + 2*1)}{9^2 + 1^2} + \frac{2*9-9*1}{9^2+1^2}i = \frac{83}{82} + \frac{9}{82}i
\]
Наращение капитала по комплексной процентной ставке. \(PV = 1\), ставка \(r = 0.09*i\)
По простой процентной ставке
\[
FV = P(1 + rt) = 1 * (1 + 9ti) = 1 + \frac{9}{100}ti
\]
package ‘ggplot2’ was built under R version 3.3.2

По сложной процентной ставке
\[
FV = P * (1 + r)^t = 1*(1 + \frac{9}{100}i)^t = (\sqrt{\frac{1082}{1000}}*(\cos(\arctan(\frac{9}{100})) + i*\sin(\arctan(\frac{9}{100}))))^t = (\frac{1082}{1000})^{\frac{t}{2}}*(\cos(t*\arctan(\frac{9}{100})) + i*sin(t*\arctan(\frac{9}{100})))
\]

По простой учетной ставке
\[
FV = \frac{P}{1-dt} = \frac{1}{1 - \frac{9t}{100}i} = \frac{1}{\sqrt{\frac{1000+82t^2}{1000}}*e^{i*\arctan(-\frac{9t}{100})}} = \sqrt{\frac{1000}{1000 + 82*t^2}}*e^{i*\arctan(\frac{9t}{100})}
\]

Графики комплексных функций
Нету технической возможности строить графики комплексных функций
Найдите сложную комплексную процентную ставку, при которой
- период наращения – 2 года,
- коэффициент наращения за этот период \(1 + 9*i\) %
Пусть процентная ставка имеем вид \[
r=a+bi
\]
Определим рост \(FV\) капитала по сложной процентной ставке, как функцию от времени: \[
FV(t) = PV(1 + r)^t = \rho(\cos\alpha + i\sin\alpha)^t = p^t(\cos\alpha t + i\sin\alpha t)
\] где \[
\rho = \sqrt{((1+a)^2 + b^2)}
\] \[
\alpha = \arctan \frac{b}{1+a}
\] Найдем точки экстремумов функции \(FV(t)\)
\[
\frac{dFV}{dt} = 0
\] \[
t_{k} = \frac{\arctan (\frac{\ln \rho}{\alpha}) + \pi k}{\alpha}
\]
Получем, следующую формулу для периода наращения
\[
T_{наращения} = \frac{\arctan (\frac{\ln \rho}{\alpha}) + 2\pi}{\alpha} - \frac{\arctan (\frac{\ln \rho}{\alpha}) }{\alpha} = \frac{2\pi}{\arctan \frac{b}{1+a}}
\] Коофиценнт наращения можно определить следующим образом \[
t_0 = \frac{\arctan (\frac{\ln \rho}{\alpha})}{\alpha}\\
k_{наращения} = \frac{FV(t_0 + T_{наращения})}{FV(t_0)} = \frac{p^{t_0+T}(\cos\alpha (t_0+T) + i\sin\alpha (t_0 + T)}{p^{t_0}(\cos\alpha t_0 + i\sin\alpha t_0)} = \frac{p^{T}(\cos\alpha (t_0+T) + i\sin\alpha (t_0 + T)}{(\cos\alpha t_0 + i\sin\alpha t_0)} = \frac{p^{\arctan \frac{b}{1+a}}(\cos\alpha (t_0+\arctan \frac{b}{1+a}) + i\sin\alpha (t_0 + \arctan \frac{b}{1+a})}{(\cos\alpha t_0 + i\sin\alpha t_0)}
\]
Итак, что бы решить задачу надо решить систему уравнений
\[
\begin{cases}
\LARGE{\frac{2\pi}{\arctan \frac{b}{1+a}} = 2 }\\
\\
\LARGE{\frac{p^{T}(\cos\alpha (t_0+T) + i\sin\alpha (t_0 + T)}{(\cos\alpha t_0 + i\sin\alpha t_0)} = 1 + 9i}
\end{cases}
\]
Графический метод нахождения комплексных корней полиномов
Задан многочлен n-порядка P(x) (n=4). Постройте на комплексной плоскости множество комплексных корней уравнения P(x)=a, где параметр а пробегает некоторый вещественный интервал с заданным шагом.
\[
x^4 + x^3 + x^3 + x^2 + x = a
\]
a_seq = seq(-3,3,by=0.005)
getRoots <- function(a) polyroot(c(a,1,1,1,1))
roots <- unlist(Map(getRoots, a_seq))
gf_data <- data.frame(x=Re(roots), y=Im(roots))
ggplot(gf_data, aes(x,y)) + geom_point(size=1) + ggtitle('Комплексные корни уравнения 4 степени')

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