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

1 Answer

3+15i3+15i

Explanation:

Given complex number

8+9i8+9i

=\sqrt145(\cos(\tan^{-1}(9/8))+i\sin(\tan^{-1}(9/8)))=145(cos(tan1(98))+isin(tan1(98)))

-5+6i5+6i

=\sqrt61(\cos(\pi-\tan^{-1}(6/5))+i\sin(\pi-\tan^{-1}(6/5)))=61(cos(πtan1(65))+isin(πtan1(65)))

=\sqrt61(-\cos(\tan^{-1}(6/5))+i\sin(\tan^{-1}(6/5)))=61(cos(tan1(65))+isin(tan1(65)))

Now, adding both the complex numbers we get

(8+9i)+(-5+6i)(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)))=145(cos(tan1(98))+isin(tan1(98)))+61(cos(tan1(65))+isin(tan1(65)))

=\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))}=145cos(tan1(98))61cos(tan1(65))+i{145sin(tan1(98))+61sin(tan1(65))}

=\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))}=145cos(cos1(8145))61cos(cos1(561))+i{145sin(sin1(9145))+61sin(tan1(661))}

=\sqrt145\(8/\sqrt145)-\sqrt61(5/\sqrt61)+i{\sqrt145(9/\sqrt145)+\sqrt61(6/\sqrt61)}=145(8145)61(561)+i{145(9145)+61(661)}

=8-5+i(9+6)=85+i(9+6)

=3+15i=3+15i