Question

12 \frac{7 + 3i}{2 + i}

Explanation:

Step1: Multiply numerator and denominator by conjugate of denominator

To simplify the complex fraction \(\frac{7 + 3i}{2 + i}\), we multiply the numerator and the denominator by the complex conjugate of the denominator, which is \(2 - i\).
\[
\frac{(7 + 3i)(2 - i)}{(2 + i)(2 - i)}
\]

Step2: Expand the numerator and the denominator

First, expand the numerator using the distributive property (FOIL method):
\[

$$\begin{align*} (7 + 3i)(2 - i)&=7\times2-7\times i+3i\times2 - 3i\times i\\ &=14-7i + 6i-3i^{2} \end{align*}$$

\]
Since \(i^{2}=- 1\), we have:
\[
14-7i + 6i-3\times(-1)=14-7i + 6i + 3=17 - i
\]
Next, expand the denominator using the difference of squares formula \((a + b)(a - b)=a^{2}-b^{2}\), where \(a = 2\) and \(b = i\):
\[
(2 + i)(2 - i)=2^{2}-i^{2}=4-(-1)=4 + 1=5
\]

Step3: Simplify the fraction

Now we have:
\[
\frac{17 - i}{5}=\frac{17}{5}-\frac{1}{5}i
\]
Or we can write it as a complex number in the form \(a+bi\) where \(a = \frac{17}{5}\) and \(b=-\frac{1}{5}\), but if we want to write it as a whole number and a fraction or in integer form (if possible), we can also check if we made a miscalculation. Wait, let's re - check the expansion of the numerator:

\[

$$\begin{align*} (7 + 3i)(2 - i)&=14-7i+6i - 3i^{2}\\ &=14 - i+3\\ &=17 - i \end{align*}$$

\]
And the denominator \((2 + i)(2 - i)=4 - i^{2}=4 + 1 = 5\). So \(\frac{17 - i}{5}=\frac{17}{5}-\frac{1}{5}i = 3.4-0.2i\). But if we want to write it in the form of a complex number with integer coefficients (by mistake in calculation? Wait, no, let's re - do the numerator expansion:

Wait, \(7\times2 = 14\), \(7\times(-i)=-7i\), \(3i\times2 = 6i\), \(3i\times(-i)=-3i^{2}\). So \(14-7i + 6i-3i^{2}=14 - i+3=17 - i\). Denominator is \(5\). So the simplified form is \(\frac{17}{5}-\frac{1}{5}i\) or we can also write it as \(\frac{17 - i}{5}\). But if we made a mistake in the problem, maybe we want to get a real and imaginary part with integers. Wait, let's check again:

Wait, maybe I made a mistake in the multiplication. Let's re - calculate the numerator:

\((7 + 3i)(2 - i)=7\times2+7\times(-i)+3i\times2+3i\times(-i)=14-7i + 6i-3i^{2}=14 - i + 3=17 - i\). Denominator: \((2 + i)(2 - i)=4 - i^{2}=4 + 1 = 5\). So the result is \(\frac{17 - i}{5}=3.4-0.2i\). But if we want to express it as a complex number in the form \(a + bi\) where \(a\) and \(b\) are fractions, it is \(\frac{17}{5}-\frac{1}{5}i\). Alternatively, if we consider that maybe there was a miscalculation and we should get an integer result, let's check the original problem again. The original fraction is \(\frac{7 + 3i}{2 + i}\). Let's try another way:

Let \(z=\frac{7 + 3i}{2 + i}\), then \(z(2 + i)=7 + 3i\), let \(z=a + bi\), then \((a + bi)(2 + i)=7 + 3i\), \(2a+ai+2bi+bi^{2}=7 + 3i\), \(2a - b+(a + 2b)i=7 + 3i\). Then we have the system of equations:
\(

$$\begin{cases}2a - b=7\\a + 2b=3\end{cases}$$

\)
From the first equation, \(b = 2a - 7\). Substitute into the second equation:
\(a+2(2a - 7)=3\)
\(a + 4a-14 = 3\)
\(5a=17\)
\(a=\frac{17}{5}\)
Then \(b=2\times\frac{17}{5}-7=\frac{34}{5}-\frac{35}{5}=-\frac{1}{5}\)
So the result is \(\frac{17}{5}-\frac{1}{5}i\) or \(3.4 - 0.2i\)

Answer:

\(\frac{17}{5}-\frac{1}{5}i\) (or \(3.4 - 0.2i\))