# 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...?

Relevance
• (2 + 8i)(2 - 8i)

Difference of Squares, or use FOIL

= 2² – (8i)²

= 2² – 64 * i²

i² = -1

= 2² – 64(-1)

= 4 +64

=

• (2 + 8i)(2 - 8i)

= 2² – (8i)²

= 4 + 64

= 68

• 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.

• (2 + 8i)(2 - 8i)

= 68

• 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?

:)>

• (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]

• 2^2-64i^2

*******************

• 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)

• (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