$z=a+bi,\ (a,b \in \mathbb{R})$, ta có: $$\begin{cases} \cos \varphi = \frac{a}{\sqrt{a^2+(b+\sqrt{2})^2}} \\ \sin \varphi = \frac{b+\sqrt{2}}{\sqrt{a^2+(b+\sqrt{2})^2}} \\ \cos ( \varphi - \frac{\pi}{4} ) = \frac{a+\sqrt{2}}{\sqrt{(a+\sqrt{2})^2+b^2}} \\ \sin ( \varphi - \frac{\pi}{4} ) = \frac{b}{\sqrt{(a+\sqrt{2})^2+b^2}}\end{cases} $$
Mặt khác: $\begin{cases} \sin \varphi + \cos \varphi = \sqrt{2} \cos ( \varphi - \frac{\pi}{4}) \\ \sin \varphi - \cos \varphi = \sqrt{2} \sin ( \varphi - \frac{\pi}{4}) \end{cases} \Leftrightarrow \begin{cases} \frac{a+b+\sqrt{2}}{\sqrt{a^2+(b+\sqrt{2})^2}} = \frac{a\sqrt{2}+2}{\sqrt{(a+\sqrt{2})^2+b^2}} & (1) \\ \frac{a-b-\sqrt{2}}{\sqrt{a^2+(b+\sqrt{2})^2}} =\frac{b\sqrt{2}}{\sqrt{(a+\sqrt{2})^2+b^2}} & (2) \end{cases} \Rightarrow $
Phức tạp quá =((. Đề thi thử trường nào vậy nhỉ :-?