Question #49a01

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Answer 1 · 2017-04-04T02:51:27

Given: #(5+2i)/(3+2i)#

Please notice that the denominator is #3+2i#, this tells us to we multiply by 1 in the form of #(3-2i)/(3-2i)#:

#(5+2i)/(3+2i)(3-2i)/(3-2i)#

The reason why we chose to do this is, because it make the denominator fit the pattern #(x+y)(x-y) = x^2-y^2#.

Please look at what happens to the denominator:

#((5+2i)(3-2i))/((3+2i)(3-2i)) = ((5+2i)(3-2i))/(3^2-(2i)^2)#

See how the denominator fits the pattern?

Square the terms in the denominator:

#((5+2i)(3-2i))/(9-4i^2)#

Use the property #i^2 = -1#

#((5+2i)(3-2i))/(9-4(-1))#

That changes the minus sign in the denominator to a plus sign:

#((5+2i)(3-2i))/(9+4)#

Perform the addition:

#((5+2i)(3-2i))/13#

Use the F.O.I.L method on the numerator:

#((15-10i+6i-4i^2))/13#

Do the same thing with #i^2= -1# in the numerator:

#((15-10i+6i-4(-1)))/13#

Change the sign:

#(15-10i+6i+4)/13#

Combine like terms:

#(19-4i)/13#

You can leave it like this or you can divide each term by 13:

#19/13-4/13i#