How do you divide $\frac{i - 1}{i - 2}$ $( i-1) / (i -2 )$ in trigonometric form?

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How do you divide $\frac{i - 1}{i - 2}$ $( i-1) / (i -2 )$ in trigonometric form?

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Answer 1 · 2016-04-10T06:45:04

$C = \frac{3 - i}{3} = 1 - \frac{i}{3}$ $C = (3-i)/3 = 1-i/3$

Explanation:

Given: $C = \frac{C_{n}}{C_{d}} = \frac{- 1 + i}{- 2 + i}$ $C= C_n/C_d=( -1+i)/(-2 +i )$
Required: The resultant to $\frac{C_{n}}{C_{d}}$ $C_n/C_d$
Solution Strategy: Multiply both nominator and denominator by complex conjugate of the denominator, $C_{d} = - 2 + i$ $C_d= -2+i$ .
Complex conjugate of $C_{d}^{*} = (- 2 - i)$ $C_d^"*" =(-2 -i)$
Thus:
$C = \frac{C_{n} \circ C_{d}^{*}}{C_{d} \circ C_{d}^{*}} = \frac{(- 1 + i) \circ (- 2 - i)}{(- 2 + i) \circ (- 2 - i)}$ $C = (C_n@C_d^"*")/(C_d@C_d^"*")=[( -1+i)@(-2 -i)]/[(-2+i )@(-2 -i)]$
$C = \frac{3 - i}{3} = 1 - \frac{i}{3}$ $C = (3-i)/3 = 1-i/3$

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How do you divide $\frac{i - 1}{i - 2}$ $( i-1) / (i -2 )$ in trigonometric form?

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