datz=a+bi$taco:a+bi+\frac{1}{a+bi}=\frac{2}{5}\times \left ( 3+4i\right )⇔a2+b2+1=25×(3+4i)(a−bi)\Leftrightarrow a^{2}+b^{2}+1=\frac{2}{5}\times \left ( 3a+4b \right )+\frac{2}{5}\times \left ( 4a-3b \right )i$$\Leftrightarrow \left\{ \begin{array}{l}a^{2}+b^{2}+1=\frac{6a}{5}+\frac{8b}{5} \\ \frac{8a}{5}-\frac{6b}{5}=0 \end{array} \right.⇒[{b=−8+2√709a=−8+2√7012\left[ {} \right.\left\{ b=−8−2√709a=−8−2√7012 \right.$$\Rightarrow z$
mk nham xl moi ng