How do you divide ( -i+4) / (7i +5 )−i+47i+5 in trigonometric form?

1 Answer
Leland Adriano Alejandro
Jan 27, 2016

sqrt1258/74*[cos (tan^-1 ((-33)/13))+isin (tan^-1 ((-33)/13))]√125874⋅[cos(tan−1(−3313))+isin(tan−1(−3313))] OR

sqrt1258/74*[cos(-68.4986^@)+isin(-68.4986^@)]√125874⋅[cos(−68.4986∘)+isin(−68.4986∘)]

Explanation:

(-i+4)/(7i+5)=(4-i)/(5+7i)−i+47i+5=4−i5+7i

(4-i)/(5+7i)*(5-7i)/(5-7i)=(20-5i-28i-7)/(25+49)=(13-33i)/744−i5+7i⋅5−7i5−7i=20−5i−28i−725+49=13−33i74

13/74-33/74i1374−3374i

Compute the magnitude rr ,let x=13/74x=1374 and y=(-33)/74y=−3374

r=sqrt(x^2+y^2)=sqrt((13/74)^2+((-33)/74)^2)=sqrt1258/74r=√x2+y2=√(1374)2+(−3374)2=√125874

Compute the Argument phiϕ

phi=tan^-1 (y/x)=tan^-1 (((-33)/cancel74)/(13/cancel74))=tan^-1 ((-33)/13)

phi=-68.4986^@

so that

(-i+4)/(7i+5)=(4-i)/(5+7i)

(4-i)/(5+7i)=sqrt1258/74*[cos(-68.4986^@)+isin(-68.4986^@)]

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