Answer:
[tex]\frac{4}{3 - i} = \frac{6}{5} + \frac{2}{5}i[/tex]
Step-by-step explanation:
Given
[tex]\frac{4}{3 - i}[/tex]
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Required
Correct Ramon's error
Start by rationalizing the expression:
[tex]\frac{4}{3 - i} = \frac{4}{3 - i}*\frac{3 + i}{3 + i}[/tex]
[tex]\frac{4}{3 - i} = \frac{4(3 + i)}{(3 - i)(3 + i)}[/tex]
Expand
[tex]\frac{4}{3 - i} = \frac{12 + 4i}{3^2 - i^2}[/tex]
[tex]\frac{4}{3 - i} = \frac{12 + 4i}{9 - i^2}[/tex]
In complex numbers:
[tex]i^2 = -1[/tex]
So:
[tex]\frac{4}{3 - i} = \frac{12 + 4i}{9 - (-1)}[/tex]
[tex]\frac{4}{3 - i} = \frac{12 + 4i}{9 +1}[/tex]
[tex]\frac{4}{3 - i} = \frac{12 + 4i}{10}[/tex]
Split fraction
[tex]\frac{4}{3 - i} = \frac{12}{10} + \frac{4i}{10}[/tex]
[tex]\frac{4}{3 - i} = \frac{6}{5} + \frac{2}{5}i[/tex]