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4

Ramon was asked to write the quotient 3 - i

in the

a + bi form. He began this way:

(3-i)

(3-i)

4(3-i)

(3² + (²)

112 - 41)

9 - 1

(12 - 4i)

8

11

(3-i)

2

Find Ramon's error and correct it (algebraically or in

words) so that he arrives at the accurate final solution.

Sagot :

MrRoyal

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]

See attachment for question

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]

View image MrRoyal
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