$dat z=a+bi$$taco:a+bi+\frac{1}{a+bi}=\frac{2}{5}\times \left ( 3+4i\right )$$\Leftrightarrow a^{2}+b^{2}+1=\frac{2}{5}\times \left ( 3+4i \right )\left ( a-bi \right )$$\Leftrightarrow a^{2}+b^{2}+1=\frac{2}{5}\times \left ( 3a+4b \right )+\frac{2}{5}\times \left ( 4a-3b \right )i$$\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.$$\Rightarrow\left[ {} \right.\left\{ \begin{array}{l} b=\frac{-8+2\sqrt{70}}{9}\\ a=\frac{-8+2\sqrt{70}}{12}\end{array} \right. $$\left[ {} \right.$ $\left\{ \begin{array}{l}b= \frac{-8-2\sqrt{70}}{9}\\ a=\frac{-8-2\sqrt{70}}{12}\end{array} \right.$$\Rightarrow z$
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