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Answer

$$\frac{3-i}{2+1}+(1+i)\cdot(1-i)\cdot(3-4i)=\frac{-25i+21}{3}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{3-i}{2+1}+(1+i)\cdot(1-i)\cdot(3-4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3-i}{3}+(1+i)\cdot(1-i)\cdot(3-4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3-i}{3}+(1-i+i-i^2)\cdot(3-4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3-i}{3}+(-i^2+1)\cdot(3-4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3-i}{3}+(1+1)\cdot(3-4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{3-i}{3}+2\cdot(3-4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{3-i}{3}+6-8i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{-25i+21}{3}\end{aligned} $$
①

Simplify denominator

$$ \color{blue}{2} + \color{blue}{1} = \color{blue}{3} $$
②

Multiply each term of $ \left( \color{blue}{1+i}\right) $ by each term in $ \left( 1-i\right) $.

$$ \left( \color{blue}{1+i}\right) \cdot \left( 1-i\right) = 1 -\cancel{i}+ \cancel{i}-i^2 $$
③

Combine like terms:

$$ 1 \, \color{blue}{ -\cancel{i}} \,+ \, \color{blue}{ \cancel{i}} \,-i^2 = -i^2+1 $$
④
$$ -i^2 = -(-1) = 1 $$
⑤

Combine like terms:

$$ \color{blue}{1} + \color{blue}{1} = \color{blue}{2} $$
⑥
Multiply $ \color{blue}{2} $ by $ \left( 3-4i\right) $ $$ \color{blue}{2} \cdot \left( 3-4i\right) = 6-8i $$
⑦

Add $ \dfrac{3-i}{3} $ and $ 6-8i $ to get $ \dfrac{ \color{purple}{ -25i+21 } }{ 3 }$.

Step 1: Write $ 6-8i $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{3}$.

$$ \begin{aligned} \frac{3-i}{3} +6-8i & \xlongequal{\text{Step 1}} \frac{3-i}{3} + \frac{6-8i}{\color{red}{1}} = \frac{ 3-i }{ 3 } + \frac{ \left( 6-8i \right) \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3-i } }{ 3 } + \frac{ \color{purple}{ 18-24i } }{ 3 }=\frac{ \color{purple}{ -25i+21 } }{ 3 } \end{aligned} $$
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