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38. \\(\\frac{-1 - 9i}{5 + 2i}\\)

Question

  1. \\(\frac{-1 - 9i}{5 + 2i}\\)

Explanation:

Step1: Multiply numerator and denominator by the conjugate of the denominator.

The conjugate of \(5 + 2i\) is \(5 - 2i\). So we multiply the fraction \(\frac{-1 - 9i}{5 + 2i}\) by \(\frac{5 - 2i}{5 - 2i}\).

$$ \frac{(-1 - 9i)(5 - 2i)}{(5 + 2i)(5 - 2i)} $$

Step2: Expand the numerator and the denominator.

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

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Since \(i^2 = -1\), we substitute that in:

$$ -5 + 2i - 45i + 18\times(-1)= -5 - 18 + 2i - 45i = -23 - 43i $$

Now, expand the denominator using the formula \((a + b)(a - b)=a^2 - b^2\):

$$ (5 + 2i)(5 - 2i)=5^2 - (2i)^2 = 25 - 4i^2 $$

Substitute \(i^2 = -1\):

$$ 25 - 4\times(-1)=25 + 4 = 29 $$

Step3: Write the fraction with the expanded numerator and denominator.

Now we have \(\frac{-23 - 43i}{29}\), which can be written as \(-\frac{23}{29}-\frac{43}{29}i\).

Answer:

\(-\frac{23}{29}-\frac{43}{29}i\)