Đáp án:
$D.\ -2^{1010}$
Giải thích các bước giải:
\(\begin{array}{l}
\quad z^2 -4z +5 =0\\
\Leftrightarrow \left[\begin{array}{l}z_1 = 2-i\\z_2 = 2+i\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}z_1 - 1 = 1 - i\\z_2 - 1 = 1 + i\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}z_1 - 1 = \sqrt2\left(\dfrac{\sqrt2}{2} - i\dfrac{\sqrt2}{2}\right)\\z_2 - 1 = \sqrt2\left(\dfrac{\sqrt2}{2} + i\dfrac{\sqrt2}{2}\right)\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}z_1 - 1 = \sqrt2\left(\cos\dfrac{\pi}{4} - i\sin\dfrac{\pi}{4}\right)\\z_2 - 1 = \sqrt2\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}(z_1 - 1)^{2019} = \left(\sqrt2\right)^{2019}\left(\cos\dfrac{\pi}{4} - i\sin\dfrac{\pi}{4}\right)^{2019}\\(z_2 - 1)^{2019} = \left(\sqrt2\right)^{2019}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)^{2019}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}(z_1 - 1)^{2019} = \left(\sqrt2\right)^{2019}\left(\cos\dfrac{2019\pi}{4} - i\sin\dfrac{2019\pi}{4}\right)\\(z_2 - 1)^{2019} = \left(\sqrt2\right)^{2019}\left(\cos\dfrac{2019\pi}{4} + i\sin\dfrac{2019\pi}{4}\right)\end{array}\right.\qquad \text{(Công thức $Moivre$)}\\
\Leftrightarrow \left[\begin{array}{l}(z_1 - 1)^{2019} = \left(\sqrt2\right)^{2019}\left(-\dfrac{\sqrt2}{2} - i\dfrac{\sqrt2}{2}\right)\\(z_2 - 1)^{2019} = \left(\sqrt2\right)^{2019}\left(-\dfrac{\sqrt2}{2} + i\dfrac{\sqrt2}{2}\right)\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}(z_1 - 1)^{2019} = -\dfrac{\left(\sqrt2\right)^{2020}}{2} - i\dfrac{\left(\sqrt2\right)^{2020}}{2}\\(z_2 - 1)^{2019} = -\dfrac{\left(\sqrt2\right)^{2020}}{2} +i\dfrac{\left(\sqrt2\right)^{2020}}{2}\end{array}\right.\\
\Leftrightarrow (z_1 - 1)^{2019}+(z_2 - 1)^{2019}=-\dfrac{\left(\sqrt2\right)^{2020}}{2}-\dfrac{\left(\sqrt2\right)^{2020}}{2}\\
\Leftrightarrow (z_1 - 1)^{2019}+(z_2 - 1)^{2019}=-\left(\sqrt2\right)^{2020}\\
\Leftrightarrow (z_1 - 1)^{2019}+(z_2 - 1)^{2019}=-2^{1010}
\end{array}\)
