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

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Answer 1 · 2016-01-27T08:09:43

#sqrt1258/74*[cos (tan^-1 ((-33)/13))+isin (tan^-1 ((-33)/13))]# OR

#sqrt1258/74*[cos(-68.4986^@)+isin(-68.4986^@)]#

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

#(4-i)/(5+7i)*(5-7i)/(5-7i)=(20-5i-28i-7)/(25+49)=(13-33i)/74#

#13/74-33/74i#

Compute the magnitude #r# ,let #x=13/74# and #y=(-33)/74#

#r=sqrt(x^2+y^2)=sqrt((13/74)^2+((-33)/74)^2)=sqrt1258/74#

Compute the Argument #phi#

#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^@)]#