# i dont know how to solve this. can someone tell me step by step without giving away the answer... (2+8i)(2-8i) thank you in advance...?

### 9 Answers

- JimLv 712 months ago
(2 + 8i)(2 - 8i)

Difference of Squares, or use FOIL

= 2² – (8i)²

= 2² – 64 * i²

i² = -1

= 2² – 64(-1)

= 4 +64

=

- AmyLv 712 months ago
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.

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- ?Lv 712 months ago
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?

:)>

- PinkgreenLv 712 months ago
(2+8i)(2-8i)

=

2^2-(8i)^2

=

4+64

=

68

[Note that (a+b)(a-b)=a^2-b^2;

Now a=2, b=8i. Also, (8i)^2=

(8i)(8i)=64i^2= -64 because

by definition i^2=-1]

- TomVLv 712 months ago
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)(a-ib)

(a+ib)(a-ib) = a(a-ib) + ib(a-ib)

= 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)(2-8i)

(Hint: let a = 2 and b = 8)

- billrussell42Lv 712 months ago
(2+8i)(2-8i)

I don't see how I can solve it without giving you the answer...

long way

2(2-8i) + 8i(2-8i)

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)(2-8i)

4 – (8i)²

4 + 64 = 68