How do you multiply # (-1-4i)(1-3i) # in trigonometric form?

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Answer 1 · 2017-10-29T13:26:53

#-13-i#

This is an interesting questions, but a solution based on the idea of #e^(itheta) = costheta + isintheta#

Consider #-1-4i#, the modulus is #root2 ( 1^2 + 4^2 )# = #root2 17 #
also #1-3i#, the modulus is #root2 ( 1^3 + 3^2 ) # = #root2 10#

Now consider the arguments, #arg(-1-4i)# = #arctan4 + pi # as easy to see by sketching the argand diagram

and the #arg(1-3i) # = # 2pi - arctan3 #

hence # (-1-4i)(1-3i)# = #root2 17 * root 2 10 * e^(arctan4 + pi) * e^(2pi - arctan3 ) #

= #root2 170 e^((3pi + arctan4 - arctan3)i) #

Hence when computed via calculator we get; # -13-i#

Yielding an interesting result of #cos(3pi + arctan4 - arctan3) = -13/root2 170#

What is linked to #cos(arctan(x)) = (1+x^2)^(-1/2)#