Question

Số phức nào sau đây là số thực?

\(A. z=\frac{1-2i}{3-4i} +\frac{1+2i}{3+4i} . \)

\(B. z=\frac{1+2i}{3-4i} -\frac{1-2i}{3+4i} \)

\(C. z=\frac{1-2i}{3-4i} -\frac{1+2i}{3+4i}  \)

\(D. z=\frac{1-2i}{3+4i} -\frac{1+2i}{3-4i} ..\)

Answer 1

Ta chọn câu A

Ta có:
\(z=\frac{1-2i}{3-4i} +\frac{1+2i}{3+4i} =\frac{\left(1-2i\right)\left(3+4i\right)+\left(1+2i\right)\left(3-4i\right)}{\left(3-4i\right)\left(3+4i\right)} =\frac{\left(11-2i\right)+\left(11+2i\right)}{25} =\frac{22}{25} \in {\rm R}.\)
\(z=\frac{1+2i}{3-4i} -\frac{1-2i}{3+4i} =\frac{\left(1+2i\right)\left(3+4i\right)-\left(1-2i\right)\left(3-4i\right)}{\left(3-4i\right)\left(3+4i\right)} =\frac{\left(-5+10i\right)-\left(-5-10i\right)}{25} =\frac{20}{25} i=\frac{4}{5} i.\)
\(z=\frac{1-2i}{3-4i} -\frac{1+2i}{3+4i} =\frac{\left(1-2i\right)\left(3+4i\right)-\left(1+2i\right)\left(3-4i\right)}{\left(3-4i\right)\left(3+4i\right)} =\frac{\left(11-2i\right)-\left(11+2i\right)}{\left(3-4i\right)\left(3+4i\right)} =-\frac{4}{25} i.\)
\(z=\frac{1-2i}{3+4i} -\frac{1+2i}{3-4i} =\frac{\left(1-2i\right)\left(3-4i\right)-\left(1+2i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)} =\frac{\left(-5-10i\right)-\left(-5+10i\right)}{\left(3-4i\right)\left(3+4i\right)} =-\frac{20}{25} i=-\frac{4}{5} i.\)