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

1 Answer

3+15i

Explanation:

Given complex number

8+9i

=145(cos(tan1(98))+isin(tan1(98)))

5+6i

=61(cos(πtan1(65))+isin(πtan1(65)))

=61(cos(tan1(65))+isin(tan1(65)))

Now, adding both the complex numbers we get

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

=145(cos(tan1(98))+isin(tan1(98)))+61(cos(tan1(65))+isin(tan1(65)))

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

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

=145(8145)61(561)+i{145(9145)+61(661)}

=85+i(9+6)

=3+15i