Question

Perform the indicated operation and express the result as a simplified complex number. I need help with 35

Answer 1

Given:-

$\frac{3+4i}{2-i}$

To find:-

The simplified form.

At first we take conjucate and multiply below and above.

The conjucate is,

$2+i$

So now we multiply. we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}$

Now we simplify. so we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}$

We know the value of,

$i^2=-1$

Substituting the value -1. we get,

$\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{ =}\frac{6-4+11i}{2+1} \\ \text{ =}\frac{2+11i}{3} \end{gathered}$

So now we split the term to bring it into the form a+ib. so we get,

$\frac{2+11i}{3}=\frac{2}{3}+i\frac{11}{3}$

So the required solution is,

$\frac{2}{3}+i\frac{11}{3}$