How do you divide ( 3i-1) / (8i -4 )3i−18i−4 in trigonometric form?

1 Answer
Shwetank Mauria
Mar 2, 2016

1/(2sqrt2)xxe^(i(alpha-beta)12√2×ei(α−β), where alpha=tan^-1(-3)α=tan−1(−3) and beta=tan^-1(-2)β=tan−1(−2)

Explanation:

(a+bi)(a+bi) an be written in as sqrt(a^2+b^2)e^(i(tan^-1(b/a))√a2+b2ei(tan−1(ba))

Hence, (3i-1)=(-1+3i)=sqrt(1^2+3^2)(e^(itan^-1(-3/1)))=sqrt10e^(itan^-1(-3))(3i−1)=(−1+3i)=√12+32(eitan−1(−31))=√10eitan−1(−3)

Similarly, (8i-4)=(-4+8i)=sqrt(4^2+8^2)(e^(itan^-1(-8/4)))=sqrt80e^(itan^-1(-2))(8i−4)=(−4+8i)=√42+82(eitan−1(−84))=√80eitan−1(−2)

Hence (3i-1)/(8i-4)=(sqrt10e^(itan^-1(-3)))/(sqrt80e^(itan^-1(-2)))3i−18i−4=√10eitan−1(−3)√80eitan−1(−2)

or sqrt(1/8)xxe^(i(tan^-1(-3))-(tan^-1(-2))√18×ei(tan−1(−3))−(tan−1(−2)) or

1/(2sqrt2)xxe^(i(alpha-beta)12√2×ei(α−β), where alpha=tan^-1(-3)α=tan−1(−3) and beta=tan^-1(-2)β=tan−1(−2)

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