How do you add #(1-5i)+(-2+i)# in trigonometric form?

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Answer 1 · 2016-02-21T06:21:27

#sqrt17[cos(tan^-1 ((-4)/(-1)))+i sin (tan^-1 ((-4)/(-1)))]#
#sqrt17[cos(4.4674103172578)+i sin (4.4674103172578)]" ""#Radian

#sqrt17[cos(255.96375653207^@)+i sin (255.96375653207^@)]#

First we add the complex numbers
#(1-5i)+(-2+i)=#

#=(1-2)+(-5i+i)#

#=-1-4i#

Convert to trigonometric form

for complex number #a+ib#

#a+ib=sqrt(a^2+b^2)*[cos(tan^-1(b/a))+i sin (tan^-1(b/a))]#

so we let #a=-1# and #b=-4#

#-1-4i=#
#sqrt((-1)^2+(-4)^2)*[cos(tan^-1((-4)/(-1)))+i sin (tan^-1((-4)/(-1)))]#

In the complex rectangular coordinate system, this is located at the 3rd quadrant

#sqrt17[cos(tan^-1 ((-4)/(-1)))+i sin (tan^-1 ((-4)/(-1)))]#
#sqrt17[cos(4.4674103172578)+i sin (4.4674103172578)]" ""#Radian

#sqrt17[cos(255.96375653207^@)+i sin (255.96375653207^@)]#

have a nice day !