Cho số phức z = a + bi ( a, b thuộc R) thỏa mãn phương trình Tính P = a + b

$\frac{\left(\left|z\right|-1\right)\left(1+iz\right)}{z-\frac{1}{\overline{z}}}=i\iff \frac{\left(\left|z\right|-1\right)\left(1+iz\right)\overline{z}}{z\overline{z}-1}=i\left(\left|z\right|\ne 1\right)$

$\iff \frac{\left(\left|z\right|-1\right)\left(1+iz\right)\overline{z}}{{\left|z\right|}^2-1}=i\iff \frac{\left(1+iz\right)\overline{z}}{\left|z\right|+1}=i$

$\iff \overline{z}+i{\left|z\right|}^2=i\left(\left|z\right|+1\right)\iff a-bi+\left(a^2+b^2\right)i=i\left(\sqrt{a^2+b^2}+1\right)$

$\iff a+\left(-b+a^2+b^2\right)i=i\left(\sqrt{a^2+b^2}+1\right)\iff \left\{\begin{array}{l}a=0\\ b^2-b=\left|b\right|+1\end{array}\right.$

$\iff \left\{\begin{array}{l}a=0\\ \left[\begin{array}{l}\left\{\begin{array}{l}b<0\\ b=\pm 1\left(loai\right)\end{array}\right.\\ \left\{\begin{array}{l}b>0\\ b^2-2b-1=0\end{array}\right.\end{array}\right.\end{array}\right.$$\iff \left\{\begin{array}{l}a=0\\ \left[\begin{array}{l}b=1+\sqrt{2}\left(nhan\right)\\ b=1-\sqrt{2}\left(loai\right)\end{array}\right.\end{array}\right.$.

Vậy $P=a+b=1+\sqrt{2}.$

Chọn đáp án C.