Question

КОМПЛЕКСНЫЕ ЧИСЛА z^2+(1+2i) z-3i=0 ​

Answer 1

Ответ:

$z=-\dfrac12-i\pm\dfrac12\sqrt[4]{265}\Bigg(\sin\dfrac12\b{arctg}\bigg(\dfrac{16}3\bigg)+i\cos\dfrac12\b{arctg}\bigg(\dfrac{16}3\bigg)\Bigg)$

Пошаговое объяснение:

$z^2+(1+2i)z-3i=0\\z=\dfrac{-1-2i\pm\sqrt{(1+2i)^2+12i}}{2}\\z=\dfrac{-1-2i\pm\sqrt{1+4i-4+12i}}{2}\\z=\dfrac{-1-2i\pm\sqrt{-3+16i}}{2}\\z=-\dfrac12-i\pm\dfrac12\sqrt{16i-3}$

Извлечем корень:

$z = 16i-3\\|z|=-\sqrt{16^2+3^2}=-\sqrt{256+9}=-\sqrt{265}\\\arg z=\b{arctg}\bigg(-\dfrac{16}{3}\bigg)=-\b{arctg}\bigg(\dfrac{16}3\bigg)$

тогда по формуле Мурава$\sqrt z=z^\frac12=|z|^\frac12\Bigg(\cos\dfrac12\arg z+i\sin \dfrac12\arg z\Bigg)=\sqrt[4]{265}\Bigg(\sin\dfrac12\b{arctg}\bigg(\dfrac{16}3\bigg)+i\cos\dfrac12\b{arctg}\bigg(\dfrac{16}3\bigg)\Bigg)$Запишем ответ

$z=-\dfrac12-i\pm\dfrac12\sqrt[4]{265}\Bigg(\sin\dfrac12\b{arctg}\bigg(\dfrac{16}3\bigg)+i\cos\dfrac12\b{arctg}\bigg(\dfrac{16}3\bigg)\Bigg)$