How do you add #(-8+4i)# and #(-3-3i)# in trigonometric form?

Question

1072 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-15T09:17:45

#sqrt(122)[cos(tan^-1(1/(-11)))+isin(tan^-1(1/(-11)))]" " "#OR
# sqrt(122)[cos(174.806^@)+isin(174.806^@)]" " "#

Add #(-8+4i)# and #(-3-3i)# to obtain

#-11+i#

Use the formula

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

Let #a=-11# and #b=1#

magnitude#" " " "r=sqrt(a^2+b^2)=sqrt((-11)^2+1^2)=sqrt122#

the angle #theta=tan^-1 (1/(-11))=174.806^@#

#sqrt(122)[cos(tan^-1(1/(-11)))+isin(tan^-1(1/(-11)))]" " "#OR
# sqrt(122)[cos(174.806^@)+isin(174.806^@)]" " "#

have a nice day... from the Philippines..