How do you add #(8+9i)# and #(-5+6i)# in trigonometric form?

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Answer 1 · 2018-07-28T16:19:53

#3+15i#

Given complex number

#8+9i#

#=\sqrt145(\cos(\tan^{-1}(9/8))+i\sin(\tan^{-1}(9/8)))#

#-5+6i#

#=\sqrt61(\cos(\pi-\tan^{-1}(6/5))+i\sin(\pi-\tan^{-1}(6/5)))#

#=\sqrt61(-\cos(\tan^{-1}(6/5))+i\sin(\tan^{-1}(6/5)))#

Now, adding both the complex numbers we get

#(8+9i)+(-5+6i)#

#=\sqrt145(\cos(\tan^{-1}(9/8))+i\sin(\tan^{-1}(9/8)))+\sqrt61(-\cos(\tan^{-1}(6/5))+i\sin(\tan^{-1}(6/5)))#

#=\sqrt145\cos(\tan^{-1}(9/8))-\sqrt61\cos(\tan^{-1}(6/5))+i{\sqrt145\sin(\tan^{-1}(9/8))+\sqrt61\sin(\tan^{-1}(6/5))}#

#=\sqrt145\cos(\cos^{-1}(8/\sqrt145))-\sqrt61\cos(\cos^{-1}(5/\sqrt61))+i{\sqrt145\sin(\sin^{-1}(9/\sqrt145))+\sqrt61\sin(\tan^{-1}(6/\sqrt61))}#

#=\sqrt145\(8/\sqrt145)-\sqrt61(5/\sqrt61)+i{\sqrt145(9/\sqrt145)+\sqrt61(6/\sqrt61)}#

#=8-5+i(9+6)#

#=3+15i#