Question

1. \\(\frac{-7}{5i}\\) 5. \\(\frac{1 + 4i}{2i}\\)

Explanation:

To solve the problem of simplifying complex fractions involving imaginary numbers, we can use the method of multiplying the numerator and denominator by the conjugate of the denominator to eliminate the imaginary unit from the denominator. Let's take the first problem: \(\frac{-7}{5i}\)

Step 1: Multiply numerator and denominator by \(i\)

To eliminate \(i\) from the denominator, we multiply the numerator and denominator by \(i\) (since the conjugate of \(5i\) is \(-5i\), but multiplying by \(i\) also works here as \(i \times i = -1\)):
\[
\frac{-7}{5i} \times \frac{i}{i} = \frac{-7i}{5i^2}
\]

Step 2: Simplify \(i^2\)

Recall that \(i^2 = -1\), so substitute \(i^2\) with \(-1\):
\[
\frac{-7i}{5(-1)} = \frac{-7i}{-5} = \frac{7i}{5}
\]

Now let's take the second problem: \(\frac{1 + 4i}{2i}\)

Step 1: Multiply numerator and denominator by \(i\)

Multiply the numerator and denominator by \(i\) to eliminate \(i\) from the denominator:
\[
\frac{1 + 4i}{2i} \times \frac{i}{i} = \frac{(1 + 4i)i}{2i^2}
\]

Step 2: Distribute \(i\) in the numerator

Distribute \(i\) in the numerator:
\[
\frac{i + 4i^2}{2i^2}
\]

Step 3: Simplify \(i^2\)

Again, substitute \(i^2\) with \(-1\):
\[
\frac{i + 4(-1)}{2(-1)} = \frac{i - 4}{-2}
\]

Step 4: Separate the real and imaginary parts

Separate the real and imaginary parts in the numerator:
\[
\frac{-4 + i}{-2} = \frac{-4}{-2} + \frac{i}{-2} = 2 - \frac{i}{2}
\]

Final Answers

1. \(\frac{-7}{5i} = \frac{7i}{5}\)
2. \(\frac{1 + 4i}{2i} = 2 - \frac{i}{2}\)

Answer:

To solve the problem of simplifying complex fractions involving imaginary numbers, we can use the method of multiplying the numerator and denominator by the conjugate of the denominator to eliminate the imaginary unit from the denominator. Let's take the first problem: \(\frac{-7}{5i}\)

Step 1: Multiply numerator and denominator by \(i\)

To eliminate \(i\) from the denominator, we multiply the numerator and denominator by \(i\) (since the conjugate of \(5i\) is \(-5i\), but multiplying by \(i\) also works here as \(i \times i = -1\)):
\[
\frac{-7}{5i} \times \frac{i}{i} = \frac{-7i}{5i^2}
\]

Step 2: Simplify \(i^2\)

Recall that \(i^2 = -1\), so substitute \(i^2\) with \(-1\):
\[
\frac{-7i}{5(-1)} = \frac{-7i}{-5} = \frac{7i}{5}
\]

Now let's take the second problem: \(\frac{1 + 4i}{2i}\)

Step 1: Multiply numerator and denominator by \(i\)

Multiply the numerator and denominator by \(i\) to eliminate \(i\) from the denominator:
\[
\frac{1 + 4i}{2i} \times \frac{i}{i} = \frac{(1 + 4i)i}{2i^2}
\]

Step 2: Distribute \(i\) in the numerator

Distribute \(i\) in the numerator:
\[
\frac{i + 4i^2}{2i^2}
\]

Step 3: Simplify \(i^2\)

Again, substitute \(i^2\) with \(-1\):
\[
\frac{i + 4(-1)}{2(-1)} = \frac{i - 4}{-2}
\]

Step 4: Separate the real and imaginary parts

Separate the real and imaginary parts in the numerator:
\[
\frac{-4 + i}{-2} = \frac{-4}{-2} + \frac{i}{-2} = 2 - \frac{i}{2}
\]

Final Answers

1. \(\frac{-7}{5i} = \frac{7i}{5}\)
2. \(\frac{1 + 4i}{2i} = 2 - \frac{i}{2}\)