Question

How would I simplify these expressions? (Pre-Calculus) I know that you are supposed to multiply all of these to do so, but every time I do that, I still get the wrong answer. My textbook says that these are the answers:

19: -15 + 8i
21: 8 + 6i
23: 34
25: 85
Could anyone explain how they got these answers? Mine are way off. Is the book just wrong? In #21, how would 3 times 3 equal 8 in the answer? Am I missing something?

Answer 1

Since you have the answers, I'll just show the steps on how to get there.

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Problem 19

`(1+4i)^2\\\\(1+4i)(1+4i)\\\\1*1 + 1*4i + 4i*1 + 4i*4i\\\\1 + 4i + 4i + 16i^2\\\\1 + 4i + 4i + 16(-1)\\\\1 + 4i + 4i-16\\\\(1-16) + (4i+4i)\\\\(1-16) + (4+4)i\\\\-15 + 8i\\\\`

Keep in mind that
`i = √(-1)` by definition. Squaring both sides leads to
`i^2 = -1`

In the second step, I used the idea that x^2 = x*x. Right after that, I used the FOIL rule to expand everything out.

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Problem 21

We could follow the same idea as problem 19, but I'll use a different approach.

`(A+B)^2 = A^2+2AB+B^2\\\\(3+i)^2 = 3^2 + 2*3*i + i^2\\\\(3+i)^2 = 9 + 6i - 1\\\\(3+i)^2 = 8 + 6i\\\\`

The formula on the first line is the perfect square binomial formula.

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Problem 23

The FOIL rule can be used if you want, but I'll use the difference of squares rule instead.

`(m+n)(m-n) = m^2 - n^2\\\\(3+5i)(3-5i) = (3)^2 - (5i)^2\\\\(3+5i)(3-5i) = 9 - 25i^2\\\\(3+5i)(3-5i) = 9 - 25(-1)\\\\(3+5i)(3-5i) = 9 + 25\\\\(3+5i)(3-5i) = 34\\\\`

It turns out that multiplying any complex number of the form a+bi with its conjugate a-bi will result in a purely real number (that has no imaginary part). More specifically:
`(a+bi)(a-bi) = a^2+b^2`

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Problem 25

We could use the difference of squares rule again, but I'll show a different approach. This time using the distribution rule. The FOIL rule could also be used if you wanted.

`(6+7i)(6-7i)\\\\x(6-7i)\\\\6x-7xi\\\\6(x)-7i(x)\\\\6(6+7i)-7i(6+7i)\\\\6(6)+6(7i)-7i(6)-7i(7i)\\\\36+42i-42i-49i^2\\\\36-49(-1)\\\\36+49\\\\85\\\\`

I used x = 6+7i and the substitution property to help distribute. Lines 3 and 6 are where distribution is applied.