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Question: Find the value of \(\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=\) a). \(\dfrac{24}{13}+\dfrac{10}{13}i\)...

Find the value of 1−2i2+i+4−i3+2i=\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=
a). 2413+1013i\dfrac{24}{13}+\dfrac{10}{13}i
b). 2413−1013i\dfrac{24}{13}-\dfrac{10}{13}i
c). 1013+2413i\dfrac{10}{13}+\dfrac{24}{13}i
d). 1013−2413i\dfrac{10}{13}-\dfrac{24}{13}i

Explanation

Solution

Hint: We will normally add these fractions as we do for the fractions of real numbers. It will simplify the terms and substitute i2=−1{{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, 1−2i2+i+4−i3+2i\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}
Taking LCM and multiplying the required terms to this term in the numerator.
1−2i2+i+4−i3+2i=(1−2i)(3+2i)+(4−i)(2+i)(2+i)(3+2i)\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,
1−2i2+i+4−i3+2i=[(1×3)+(1×2i)+(−2i×3)+(−2i×2i)+(4×2)+(4×i)+(−i×2)+(−i×i)](2×3)+(2×2i)+(i×3)+(i×2i)\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)}
=3+2i−6i−4i2+8+4i−2i−i26+4i+3i+2i2 =11+(−2i)+(−5i2)6+7i+2i2 \begin{aligned} & =\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{aligned}
Now, putting i2=1{{i}^{2}}=1 we get,
=11−2i−5(−1)6+7i+2(−1) =11−2i+56+7i−2 =16−2i4+7i \begin{aligned} & =\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{aligned}
Now, multiplying and dividing by the conjugate of 4+7i, that is 4-7i, we get,
=16−2i4+7i×4−7i4−7i =(16−2i)(4−7i)(4+7i)(4−7i) \begin{aligned} & =\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{aligned}
Multiplying, we get,
=(16×4)+(16×−7i)+(−2i×4)+(−2i×−7i)(4×4)+(4×−7i)+(7i×4)+(7i×−7i) =64−112i−8i+14i216−28i+28i−49i2 =64−120i+14i216−49i2 \begin{aligned} & =\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{aligned}
Putting i2−1{{i}^{2}}-1 we get,
=64−120i+14(−1)16−49(−1) =64−120i−1416+49 =50−120i65 \begin{aligned} & =\dfrac{64-120i+14\left( -1 \right)}{16-49\left( -1 \right)} \\\ & =\dfrac{64-120i-14}{16+49} \\\ & =\dfrac{50-120i}{65} \\\ \end{aligned}
Splitting the terms,
=5065−12065i=\dfrac{50}{65}-\dfrac{120}{65}i
Simplifying we get,
=1013−2413i=\dfrac{10}{13}-\dfrac{24}{13}i
Thus, 1−2i2+i+4−i3+2i=1013−2413i\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 1−2i2+i+4−i3+2i\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}
Taking the first fraction, 1−2i2+i\dfrac{1-2i}{2+i}
Multiplying and dividing by the conjugate of 2+I, which is 2-I, we get,

& \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{aligned}$$ Putting ${{i}^{2}}=-1$ , we get, $\begin{aligned} & =\dfrac{2-5i+2\left( -1 \right)}{4-\left( -1 \right)} \\\ & =\dfrac{2-5i-2}{5} \\\ & =\dfrac{-5i}{5} \\\ & =-i \\\ \end{aligned}$ 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. $\begin{aligned} & \dfrac{4-i}{3+2i}\times \dfrac{3-2i}{3-2i} \\\ & =\dfrac{\left( 4-i \right)\left( 3-2i \right)}{\left( 3+2i \right)\left( 3-2i \right)} \\\ \end{aligned}$ Multiplying we get, $\begin{aligned} & =\dfrac{\left( 4\times 3 \right)+\left( 4\times -2i \right)+\left( -i\times 3 \right)+\left( -i\times -2i \right)}{\left( 3\times 3 \right)+\left( 3\times -2i \right)+\left( 2i\times 3 \right)+\left( 2i\times 2i \right)} \\\ & =\dfrac{12-8i-3i+2{{i}^{2}}}{9-6i+6i-4{{i}^{2}}} \\\ \end{aligned}$ Putting ${{i}^{2}}=-1$, we get, $\begin{aligned} & =\dfrac{12-11i+2\left( -1 \right)}{i-4\left( -1 \right)} \\\ & =\dfrac{12-11i-2}{9+4} \\\ & =\dfrac{10-11i}{13} \\\ \end{aligned}$ Now, adding both, $\begin{aligned} & \dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i} \\\ & =-i+\dfrac{10-11i}{13} \\\ \end{aligned}$ Taking LCM and multiplying required terms in numerator, $\begin{aligned} & =\dfrac{-13i+10-11i}{13} \\\ & =\dfrac{10-24i}{13} \\\ \end{aligned}$ Splitting the terms, $=\dfrac{10}{13}-\dfrac{24}{13}i$ .