Question

1. \\(\frac{1 + 8i}{2 - 4i}\\) 14. \\(\frac{-3 + 7i}{-9 + i}\\)

Explanation:

Problem 13: Simplify \(\frac{1 + 8i}{2 - 4i}\)

Step 1: Multiply numerator and denominator by the conjugate of the denominator.

The conjugate of \(2 - 4i\) is \(2 + 4i\). So we have:
\[
\frac{(1 + 8i)(2 + 4i)}{(2 - 4i)(2 + 4i)}
\]

Step 2: Expand the numerator and the denominator.

• Numerator: \((1 + 8i)(2 + 4i)=1\times2 + 1\times4i + 8i\times2 + 8i\times4i = 2 + 4i + 16i + 32i^{2}\)

Since \(i^{2}=-1\), this becomes \(2 + 20i + 32\times(-1)=2 + 20i - 32=-30 + 20i\)

• Denominator: \((2 - 4i)(2 + 4i)=2^{2}-(4i)^{2}=4 - 16i^{2}=4 - 16\times(-1)=4 + 16 = 20\)

Step 3: Simplify the fraction.

\[
\frac{-30 + 20i}{20}=\frac{-30}{20}+\frac{20i}{20}=-\frac{3}{2}+i
\]

Step 1: Multiply numerator and denominator by the conjugate of the denominator.

The conjugate of \(-9 + i\) is \(-9 - i\). So we have:
\[
\frac{(-3 + 7i)(-9 - i)}{(-9 + i)(-9 - i)}
\]

Step 2: Expand the numerator and the denominator.

• Numerator: \((-3 + 7i)(-9 - i)=(-3)\times(-9)+(-3)\times(-i)+7i\times(-9)+7i\times(-i)=27 + 3i - 63i - 7i^{2}\)

Since \(i^{2}=-1\), this becomes \(27 - 60i - 7\times(-1)=27 - 60i + 7 = 34 - 60i\)

• Denominator: \((-9 + i)(-9 - i)=(-9)^{2}-i^{2}=81 - (-1)=81 + 1 = 82\)

Step 3: Simplify the fraction.

\[
\frac{34 - 60i}{82}=\frac{34}{82}-\frac{60i}{82}=\frac{17}{41}-\frac{30}{41}i
\]

Answer:

\(-\frac{3}{2}+i\)

Problem 14: Simplify \(\frac{-3 + 7i}{-9 + i}\)