Yahoo Answers: Answers and Comments for I dont know how to solve this. can someone tell me step by step without giving away the answer... (2+8i)(28i) thank you in advance...? [Mathematics]
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From kendra
enGB
Fri, 23 Aug 2019 19:05:17 +0000
3
Yahoo Answers: Answers and Comments for I dont know how to solve this. can someone tell me step by step without giving away the answer... (2+8i)(28i) thank you in advance...? [Mathematics]
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From billrussell42: (2+8i)(28i)
I don't see how I can solve ...
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Fri, 23 Aug 2019 19:12:45 +0000
(2+8i)(28i)
I don't see how I can solve it without giving you the answer...
long way
2(28i) + 8i(28i)
4 – 16i + 16i – 64i²
note: 1² = –1
4 – 64(–1)
4 + 64
68
fast way, using identity a² – b² = (a + b)(a – b)
(2+8i)(28i)
4 – (8i)²
4 + 64 = 68

From Mike G: 2^264i^2
*******************
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Fri, 23 Aug 2019 21:56:27 +0000
2^264i^2
*******************

From Jim: (2 + 8i)(2  8i)
Difference of Squares, or use...
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Mon, 26 Aug 2019 21:33:18 +0000
(2 + 8i)(2  8i)
Difference of Squares, or use FOIL
= 2² – (8i)²
= 2² – 64 * i²
i² = 1
= 2² – 64(1)
= 4 +64
=

From Krishnamurthy: (2 + 8i)(2  8i)
= 2² – (8i)²
= 4 + 64
= 68
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Mon, 26 Aug 2019 11:50:42 +0000
(2 + 8i)(2  8i)
= 2² – (8i)²
= 4 + 64
= 68

From Amy: Remember FOIL? If not, just apply the distribu...
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Mon, 26 Aug 2019 01:26:25 +0000
Remember FOIL? If not, just apply the distributive property twice:
(a + b)(c + d) = (a+b)c + (a+b)d = ac + bc + ad + bd
When you apply this to the complex numbers, you will get a combination of real and imaginary terms. Remember that i^2 = 1, so the term with i^2 becomes a real number. Add the real terms together, and add the imaginary terms together.

From sepia: (2 + 8i)(2  8i)
= 68
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Sat, 24 Aug 2019 10:31:48 +0000
(2 + 8i)(2  8i)
= 68

From Wayne DeguMan: Expanding gives:
4  16i + 16i  64i²
With i...
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Sat, 24 Aug 2019 09:17:42 +0000
Expanding gives:
4  16i + 16i  64i²
With i² = 1 we have:
4  16i + 16i  64(1)
i.e. 4  16i + 16i + 64
You can complete this?
:)>

From Pinkgreen: (2+8i)(28i)
=
2^2(8i)^2
=
4+64
=
68
[Note t...
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Sat, 24 Aug 2019 02:43:31 +0000
(2+8i)(28i)
=
2^2(8i)^2
=
4+64
=
68
[Note that (a+b)(ab)=a^2b^2;
Now a=2, b=8i. Also, (8i)^2=
(8i)(8i)=64i^2= 64 because
by definition i^2=1]

From TomV: Multiply each of the terms within the second s...
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Fri, 23 Aug 2019 19:37:19 +0000
Multiply each of the terms within the second set of parentheses by each term within the first set of parentheses and combine like terms of the result.
For a general example consider (a+ib)(aib)
(a+ib)(aib) = a(aib) + ib(aib)
= a(a)  a(ib) + ib(a)  ib(ib)
= a²  aib + aib  i²b²
= a²  i²b²
Now consider that if i = √1, then i² = 1, and the summation becomes:
= a²  (1)b²
= a² + b²
Apply that general process to your specific problem to obtain the answer to (2+8i)(28i)
(Hint: let a = 2 and b = 8)