1. ## complex numbers

If z = 3x + 4i and w = 1 - 2i, calculate:

(z - w)^2

2. $\displaystyle z-w=3x-1+6i$

$\displaystyle (a+bi)^2=a^2-b^2+2abi$

Replace a with 3x-1 and b with 6.

3. Originally Posted by Joker37
If z = 3x + 4i and w = 1 - 2i, calculate:

(z - w)^2
$\displaystyle z-w = (3x+4i) - (1 - 2i) = (3x-1) + 6i$

$\displaystyle [(3x-1) + 6i]^2$

expand and simplify

4. Originally Posted by skeeter
$\displaystyle z-w = (3x+4i) - (1 - 2i) = (3x-1) + 6i$

$\displaystyle [(3x-1) + 6i]^2$

expand and simplify
(3x - 1 - 6i)(3x - 1 + 6i)?

5. Originally Posted by Joker37
(3x - 1 - 6i)(3x - 1 + 6i)?
No. (A + B)^2 = (A + B)(A + B). NOT (A + B)(A - B).

So (3x - 1 - 6i)^2 = (3x - 1 - 6i)(3x - 1 - 6i).

Virtually the same idea as here: http://www.mathhelpforum.com/math-he...x-numbers.html