Question

in the complex numbers, where $i^{2}=-1$, $\frac{2 - i}{-3 + i}=$?
a. $-\frac{2}{3}-i$
b. $-\frac{5}{8}+\frac{1}{8}i$
c. $-\frac{7}{8}+\frac{1}{8}i$
d. $-\frac{5}{10}+\frac{1}{10}i$
e. $-\frac{7}{10}+\frac{1}{10}i$

Explanation:

Step1: Multiply by conjugate

Multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $-3 + i$ is $-3 - i$. So we have $\frac{(2 - i)(-3 - i)}{(-3 + i)(-3 - i)}$.

Step2: Expand numerator

Expand $(2 - i)(-3 - i)$ using FOIL method:
\[

$$\begin{align*} (2 - i)(-3 - i)&=2\times(-3)+2\times(-i)-i\times(-3)-i\times(-i)\\ &=-6-2i + 3i+i^{2}\\ &=-6 + i- 1\\ &=-7 + i \end{align*}$$

\]

Step3: Expand denominator

Expand $(-3 + i)(-3 - i)$ using the difference - of - squares formula $(a + b)(a - b)=a^{2}-b^{2}$. Here $a=-3$ and $b = i$, so $(-3 + i)(-3 - i)=(-3)^{2}-i^{2}=9-(-1)=10$.

Step4: Simplify fraction

The fraction $\frac{-7 + i}{10}=-\frac{7}{10}+\frac{1}{10}i$.

Answer:

E. $-\frac{7}{10}+\frac{1}{10}i$