How do you multiply # (3+5i)(6-6i) # in trigonometric form?

Question

1502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T17:23:36

#6(8+2i)#

Given that

#(3+5i)(6-6i)#

#=\sqrt34(\cos(\tan^{-1}(5/3))+i\sin(\tan^{-1}(5/3)))\cdot 6\sqrt2(\cos(-\pi/4)+i\sin(-\pi/4))#

#=12\sqrt17(\cos(\tan^{-1}(5/3)-\pi/4)+i\sin(\tan^{-1}(5/3)-\pi/4))#

#=12\sqrt17(\cos(\tan^{-1}(5/3))\cos(\pi/4)+\sin(\tan^{-1}(5/3))\sin(\pi/4)+i(\sin(\tan^{-1}(5/3))\cos(\pi/4)-\cos(\tan^{-1}(5/3))\sin(\pi/4)))#

#=12\sqrt17(\frac{3}{\sqrt34}1/\sqrt2+5/\sqrt34 1/\sqrt2+i(\frac{5}{\sqrt34}1/\sqrt2-3/\sqrt34 1/\sqrt2))#

#={12\sqrt17}/{2\sqrt17}(8+2i)#

#=6(8+2i)#