If z = 3x + 4i and w = 1 - 2i, calculate: (z - w)^2
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$\displaystyle z-w=3x-1+6i$ $\displaystyle (a+bi)^2=a^2-b^2+2abi$ Replace a with 3x-1 and b with 6.
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
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)?
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
Last edited by mr fantastic; Mar 3rd 2009 at 01:56 AM.
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