Question

question 1 of 10
divide the following complex numbers:
\frac{(2+i)}{(1-4i)}

a. \frac{2}{15} - \frac{9}{15}i
b. -\frac{2}{17} + \frac{9}{17}i
c. \frac{6}{17} + \frac{9}{17}i
d. -\frac{6}{15} - \frac{9}{15}i

Explanation:

Step1: Multiply numerator and denominator by conjugate

To divide complex numbers, we multiply the numerator and denominator by the conjugate of the denominator. The conjugate of \(1 - 4i\) is \(1 + 4i\). So we have:
\[
\frac{(2 + i)(1 + 4i)}{(1 - 4i)(1 + 4i)}
\]

Step2: Expand the numerator and denominator

First, expand the numerator using the distributive property (FOIL method):
\[
(2 + i)(1 + 4i)=2\times1 + 2\times4i + i\times1 + i\times4i = 2 + 8i + i + 4i^{2}
\]
Since \(i^{2}=-1\), this becomes:
\[
2 + 9i + 4\times(-1)=2 + 9i - 4=-2 + 9i
\]
Now, expand the denominator using the difference of squares formula \((a - b)(a + b)=a^{2}-b^{2}\):
\[
(1 - 4i)(1 + 4i)=1^{2}-(4i)^{2}=1 - 16i^{2}
\]
Again, using \(i^{2}=-1\), we get:
\[
1 - 16\times(-1)=1 + 16 = 17
\]

Step3: Form the new complex number

Now we have \(\frac{-2 + 9i}{17}\), which can be written as:
\[
-\frac{2}{17}+\frac{9}{17}i
\]

Answer:

B. \(-\frac{2}{17}+\frac{9}{17}i\)