To solve the exercise you can apply the complex arithmetic rule, that is
[tex]\frac{a+bi}{c+di}=\frac{(c-di)(a+bi)}{(c-di)(c+di)}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}[/tex]
So, in this case, you have
[tex]\begin{gathered} \frac{6-i}{4+3i} \\ a=6 \\ b=-1 \\ c=4 \\ d=3 \end{gathered}[/tex][tex]\frac{6-i}{4+3i}=\frac{(6\cdot4+(-1)\cdot3)+(-1\cdot4-6\cdot3)i}{4^2+3^2}[/tex]
Simplifying
[tex]\begin{gathered} \frac{6-i}{4+3i}=\frac{(24-3)+(-4-18)i}{16+9} \\ \frac{6-i}{4+3i}=\frac{21-22i}{16+9} \\ \frac{6-i}{4+3i}=\frac{21-22i}{25} \end{gathered}[/tex]
Finally, rewrite in binomic form, that is, z = a+bi
[tex]\frac{21-22i}{25}=\frac{21}{25}-\frac{22i}{25}[/tex]
Therefore, the correct answer is B.
[tex]\frac{21}{25}-\frac{22i}{25}[/tex]
