InterviewSolution
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InterviewSolution
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Express the following in `a + ib` form: (a) `((cos alpha + i sin alpha)^(4))/((sin beta + i cos beta)^(5))` (b) `((1+ cos phi + i sin phi)/(1 + cos phi - isin phi))^(n)` (c) `((cos alpha + i sin alpha)(cos beta + i sin beta))/((cos gamma + i sin gamma)(cos delta + i sin delta))` |
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Answer» Correct Answer - (a) ` sin(4 alpha + 5beta) - i cos ( 4alpha + 5 beta)` (b) ` cos nphi+ isin nphi ` (c) `cos (alpha + beta -gamma -delta) + i sin (alpha + beta- gamma - delta)` `((cos alpha+sin alpha)^(4))/((sin beta +icos beta)^(5)) = (cos4alpha + isin 4alpha)/(i^(5)(cos beta -isin beta)^(5))` `=-i(cos 4alpha + isin 4alpha)(cos beta -isin beta)^(5)` ` =- i[cos 4alpha + isin 4alpha][cos 5 beta + isin 5 beta]` `= -i[cos(4alpha + 5beta)+isin(4alpha + 5beta)]` `= sin (4alpha + 5beta)-icos (4alpha + 5beta)` (b) `((1+cos phi + isin phi)/(1+cos phi -isin phi))^(n)` `=[(2cos^(2) (phi//2) + isin (phi//2) cos(pi//2))/(2cos^(2)(phi//2) -2isin(phi//2) cos(phi//2))]^(n)` `= [(cos(phi//2) +isin (phi//2) )/(cos(phi//2) -isin (phi//2))]^(n)` `=[((cos phi+ sin phi)^(1//2))/((cos phi +isin phi)^(-1//2))]^(n)` `=[cos phi + isin phi]^(n)` `= cos n phi + isinnphi` |
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