Multiply the vectors. The answer is 29 - j29?

Question

#(7 + j3)(2 - j5)#

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Answer 1 · 2016-11-30T02:22:27

Please see the explanation.

Use the F.O.I.L. method.

Start with the product of the F irst terms:

#(7 + j3)(2 - j5) = 14#

Add the product of the O utside terms:

#(7 + j3)(2 - j5) = 14 - j35#

Add the product of the I nside terms:

#(7 + j3)(2 - j5) = 14 - j35 + j6#

Add the product of the L ast terms:

#(7 + j3)(2 - j5) = 14 - j35 + j6 - j^(2)15#

Because #j^2 = -1#, the last term becomes + 15:

#(7 + j3)(2 - j5) = 14 - j35 + j6 + 15#

Combine like terms:

#(7 + j3)(2 - j5) = 29 - j29#