Question

Find the value of $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=$
a). $\dfrac{24}{13}+\dfrac{10}{13}i$
b). $\dfrac{24}{13}-\dfrac{10}{13}i$
c). $\dfrac{10}{13}+\dfrac{24}{13}i$
d). $\dfrac{10}{13}-\dfrac{24}{13}i$

Answer 1

Hint: We will normally add these fractions as we do for the fractions of real numbers. It will simplify the terms and substitute ${{i}^{2}}=-1$ . If there is any linear term in the denominator we will multiply and divide by its conjugate to simplify it further.

Complete step-by-step solution -
Given fractions, $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}$
Taking LCM and multiplying the required terms to this term in the numerator.
$\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=\dfrac{\left( 1-2i \right)\left( 3+2i \right)+\left( 4-i \right)\left( 2+i \right)}{\left( 2+i \right)\left( 3+2i \right)}$
Now multiplying using normal rules of multiplication, we get,
$\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=\dfrac{\left[ \left( 1\times 3 \right)+\left( 1\times 2i \right)+\left( -2i\times 3 \right)+\left( -2i\times 2i \right)+\left( 4\times 2 \right)+\left( 4\times i \right)+\left( -i\times 2 \right)+\left( -i\times i \right) \right]}{\left( 2\times 3 \right)+\left( 2\times 2i \right)+\left( i\times 3 \right)+\left( i\times 2i \right)}$
$\begin{align}
  & =\dfrac{3+2i-6i-4{{i}^{2}}+8+4i-2i-{{i}^{2}}}{6+4i+3i+2{{i}^{2}}} \\
 & =\dfrac{11+\left( -2i \right)+\left( -5{{i}^{2}} \right)}{6+7i+2{{i}^{2}}} \\
\end{align}$
Now, putting ${{i}^{2}}=1$ we get,
$\begin{align}
  & =\dfrac{11-2i-5\left( -1 \right)}{6+7i+2\left( -1 \right)} \\
 & =\dfrac{11-2i+5}{6+7i-2} \\
 & =\dfrac{16-2i}{4+7i} \\
\end{align}$
Now, multiplying and dividing by the conjugate of 4+7i, that is 4-7i, we get,
$\begin{align}
  & =\dfrac{16-2i}{4+7i}\times \dfrac{4-7i}{4-7i} \\
 & =\dfrac{\left( 16-2i \right)\left( 4-7i \right)}{\left( 4+7i \right)\left( 4-7i \right)} \\
\end{align}$
Multiplying, we get,
$\begin{align}
  & =\dfrac{\left( 16\times 4 \right)+\left( 16\times -7i \right)+\left( -2i\times 4 \right)+\left( -2i\times -7i \right)}{\left( 4\times 4 \right)+\left( 4\times -7i \right)+\left( 7i\times 4 \right)+\left( 7i\times -7i \right)} \\
 & =\dfrac{64-112i-8i+14{{i}^{2}}}{16-28i+28i-49{{i}^{2}}} \\
 & =\dfrac{64-120i+14{{i}^{2}}}{16-49{{i}^{2}}} \\
\end{align}$
Putting ${{i}^{2}}-1$ we get,
$\begin{align}
  & =\dfrac{64-120i+14\left( -1 \right)}{16-49\left( -1 \right)} \\
 & =\dfrac{64-120i-14}{16+49} \\
 & =\dfrac{50-120i}{65} \\
\end{align}$
Splitting the terms,
$=\dfrac{50}{65}-\dfrac{120}{65}i$
Simplifying we get,
$=\dfrac{10}{13}-\dfrac{24}{13}i$
Thus, $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=\dfrac{10}{13}-\dfrac{24}{13}i$ .
Therefore, option (d) is correct.

Note: This problem can also be solved by simplifying each fraction first and then adding them.
Given $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}$
Taking the first fraction, $\dfrac{1-2i}{2+i}$
Multiplying and dividing by the conjugate of 2+I, which is 2-I, we get,
\[\begin{align}
  & \dfrac{1-2i}{2+i}\times \dfrac{2-i}{2-i} \\
 & =\dfrac{\left( 1-2i \right)\left( 2-i \right)}{\left( 2+i \right)\left( 2-i \right)} \\
 & =\dfrac{\left( 1\times 2 \right)+\left( 1\times -i \right)+\left( -2i\times 2 \right)+\left( -2i\times -i \right)}{\left( 2\times 2 \right)+\left( 2\times -i \right)+\left( i\times 2 \right)+\left( i\times -i \right)} \\
 & =\dfrac{2-i-4i+2{{i}^{2}}}{4-2i+2i-{{i}^{2}}} \\
\end{align}\]
Putting ${{i}^{2}}=-1$ , we get,
$\begin{align}
  & =\dfrac{2-5i+2\left( -1 \right)}{4-\left( -1 \right)} \\
 & =\dfrac{2-5i-2}{5} \\
 & =\dfrac{-5i}{5} \\
 & =-i \\
\end{align}$
Now, take the second fraction, $\dfrac{4-i}{3+2i}$
Multiplying and dividing by the conjugate of 3+2i, that is 3-2i, we get.
$\begin{align}
  & \dfrac{4-i}{3+2i}\times \dfrac{3-2i}{3-2i} \\
 & =\dfrac{\left( 4-i \right)\left( 3-2i \right)}{\left( 3+2i \right)\left( 3-2i \right)} \\
\end{align}$
Multiplying we get,
$\begin{align}
  & =\dfrac{\left( 4\times 3 \right)+\left( 4\times -2i \right)+\left( -i\times 3 \right)+\left( -i\times -2i \right)}{\left( 3\times 3 \right)+\left( 3\times -2i \right)+\left( 2i\times 3 \right)+\left( 2i\times 2i \right)} \\
 & =\dfrac{12-8i-3i+2{{i}^{2}}}{9-6i+6i-4{{i}^{2}}} \\
\end{align}$
Putting ${{i}^{2}}=-1$, we get,
$\begin{align}
  & =\dfrac{12-11i+2\left( -1 \right)}{i-4\left( -1 \right)} \\
 & =\dfrac{12-11i-2}{9+4} \\
 & =\dfrac{10-11i}{13} \\
\end{align}$
Now, adding both,
$\begin{align}
  & \dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i} \\
 & =-i+\dfrac{10-11i}{13} \\
\end{align}$
Taking LCM and multiplying required terms in numerator,
$\begin{align}
  & =\dfrac{-13i+10-11i}{13} \\
 & =\dfrac{10-24i}{13} \\
\end{align}$
Splitting the terms,
$=\dfrac{10}{13}-\dfrac{24}{13}i$ .