Given:-
[tex]\frac{3+4i}{2-i}[/tex]
To find:-
The simplified form.
At first we take conjucate and multiply below and above.
The conjucate is,
[tex]2+i[/tex]
So now we multiply. we get,
[tex]\frac{3+4i}{2-i}\times\frac{2+i}{2+i}[/tex]
Now we simplify. so we get,
[tex]\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}[/tex]
We know the value of,
[tex]i^2=-1[/tex]
Substituting the value -1. we get,
[tex]\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{ =}\frac{6-4+11i}{2+1} \\ \text{ =}\frac{2+11i}{3} \end{gathered}[/tex]
So now we split the term to bring it into the form a+ib. so we get,
[tex]\frac{2+11i}{3}=\frac{2}{3}+i\frac{11}{3}[/tex]
So the required solution is,
[tex]\frac{2}{3}+i\frac{11}{3}[/tex]
