How do you divide $\frac{i + 8}{5 i + 5}$ $( i+8) / (5i +5 )$ in trigonometric form?

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How do you divide $\frac{i + 8}{5 i + 5}$ $( i+8) / (5i +5 )$ in trigonometric form?

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Answer 1 · 2016-03-19T04:14:20

$C_{3} = \frac{40 + 5 i - 40 i + 5}{25 + 25} = \frac{45 - 35 i}{50} = \frac{1}{10} (9 - 7 i)$ $C_3 = (40+5i-40i+5)/(25+25)= (45-35i)/50= 1/10(9-7i)$

Explanation:

Given the complex set $C_{1} = (i + 8)$ $C_1 =( i+8)$ and $C_{2} = (5 i + 5)$ $C_2=(5i +5 )$
Required: $\frac{i + 8}{5 i + 5}$ $( i+8)/(5i +5 )$
Solution: Use complex conjugate of $C_{2}$ $C_2$ , $¯ ¯¯ ¯ C_{2}$ $bar(C_2)$ to perform complex number division.
The product of a complex number $C$ $C$ with it's conjugate $¯ ¯ ¯ C$ $bar(C)$ is: $R = C \cdot ¯ ¯ ¯ C = (a + b i) (a - b i) = a^{2} + b^{2}$ $R = C*bar(C) = (a+bi)(a-bi) =a^2+b^2$ , a real number.
Thus multiplying top and bottom by $¯ ¯¯ ¯ C_{2} = 5 - 5 i$ $bar(C_2) = 5-5i$
$C_{3} = \frac{C_{1} \cdot ¯ ¯¯ ¯ C_{2}}{C_{2} \cdot ¯ ¯¯ ¯ C_{2}} = \frac{(i + 8) (5 - 5 i)}{(5 + 5 i) \cdot (5 - 5 i)}$ $C_3 = (C_1*bar(C_2))/(C_2*bar(C_2)) = ((i+8)(5-5i))/((5+5i)*(5-5i))$

$C_{3} = \frac{40 + 5 i - 40 i + 5}{25 + 25} = \frac{45 - 35 i}{50} = \frac{1}{10} (9 - 7 i)$ $C_3 = (40+5i-40i+5)/(25+25)= (45-35i)/50= 1/10(9-7i)$

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How do you divide $\frac{i + 8}{5 i + 5}$ $( i+8) / (5i +5 )$ in trigonometric form?

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