Question

помогите пожалуйста.​

Answer 1

$|1+i|=|1-i|=\sqrt{1^2+1^2}=\sqrt{2};\ \arg(1+i)=\dfrac{\pi}{4};\ \arg(1-i)=-\dfrac{\pi}{4};$

$1+i=\sqrt{2}e^{i\frac{\pi}{4}}=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4});\ 1-i=\sqrt{2}e^{-i\frac{\pi}{4}}=\sqrt{2}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}));$

$(1+i)^{28}=(\sqrt{2})^{28}e^{28i\frac{\pi}{4}}=2^{14}e^{7i\pi}=2^{14}(\cos 7\pi+i\sin 7\pi)=-2^{14};$  

$(1+i)^{24}=(\sqrt{2})^{24}e^{24i\frac{\pi}{4}}=2^{12}e^{6i\pi}=2^{12}; (1-i)^{24}=2^{12}e^{-6i\pi}=2^{12};$

$\dfrac{(1+i)^{28}}{(1-i)^{24}-i(1+i)^{24}}=\dfrac{-2^{14}}{2^{12}(1-i)}=-\dfrac{2^2(1+i)}{(1-i)(1+i)}=-\dfrac{4(1+i)}{2}=-2-2i.$

Ответ: -2-2i