Вычислите (2+i)^7+(2-i)^7

Question

Дан 1 ответ

Правильный ответ

0\; ,\; \; sin\alpha =\frac{1}{\sqrt5}>0\; ,\; \; tg\alpha =\frac{1}{2}\; ,\; \; \alpha =arctg\frac{1}{2}\; ,\; argz=\alpha \\\\2+i=\sqrt5\cdot \Big (cos(arctg\frac{1}{2})+i\, sin(arctg\frac{1}{2})\Big )\\\\(2+i)^7=\sqrt{5^7}\cdot \Big (cos(7arctg\frac{1}{2})+i\, sin(7arctg\frac{1}{2})\Big )\\\\\overline {z}=2-i=\sqrt5\cdot \Big (cos(-arctg\frac{1}{2})+i\, sin(-arctg\frac{1}{2})\Big )" alt="(2+i)^7+(2-i)^7=?\\\\z=2+i\; ,\; \; |z|=r=\sqrt{2^2+1}=\sqrt5\; ,\\\\cos\alpha =\frac{2}{\sqrt5}>0\; ,\; \; sin\alpha =\frac{1}{\sqrt5}>0\; ,\; \; tg\alpha =\frac{1}{2}\; ,\; \; \alpha =arctg\frac{1}{2}\; ,\; argz=\alpha \\\\2+i=\sqrt5\cdot \Big (cos(arctg\frac{1}{2})+i\, sin(arctg\frac{1}{2})\Big )\\\\(2+i)^7=\sqrt{5^7}\cdot \Big (cos(7arctg\frac{1}{2})+i\, sin(7arctg\frac{1}{2})\Big )\\\\\overline {z}=2-i=\sqrt5\cdot \Big (cos(-arctg\frac{1}{2})+i\, sin(-arctg\frac{1}{2})\Big )" align="absmiddle" class="latex-formula">

$(2-i)^7=\sqrt{5^7}\cdot \Big (cos(-7arctg\frac{1}{2})+i\, sin(-7arctg\frac{1}{2})\Big )\\\\(2+i)^7+(2-i)^7=\sqrt{5^7}\cdot \Big (cos(7arctg\frac{1}{2})+i\, sin(7arctg\frac{1}{2})\Big ) +\\\\+\sqrt{5^7}\cdot \Big (cos(-7arctg\frac{1}{2})+i\, sin(-7arctg\frac{1}{2})\Big )$

$(2+i)^7+(2-i)^7=(2^7+7\cdot 2^6\cdot i+21\cdot 2^5\cdot i^2+35\cdot 2^4\cdot i^3+35\cdot 2^3\cdot i^4+\\\\+21\cdot 2^2\cdot i^5+7\cdot 2\cdot i^6+i^7)+(2^7-7\cdot 2^6\cdot i+21\cdot 2^5\cdot i^2-35\cdot 2^4\cdot i^3+\\\\+35\cdot 2^3\cdot i^4-21\cdot 2^2\cdot i^5+7\cdot 2\cdot i^6-i^7)=\\\\=2\cdot (2^7+21\cdot 2^5\cdot i^2+35\cdot 2^3\cdot i^4+7\cdot 2\cdot i^6)=\\\\=2\cdot (128-672+280-14)=-556$

NNNLLL54_zn (834k баллов) 30 Май, 20