# Complex Number Kookiness 6

## (1 + *i*)(x + x*i*) = 2x*i*

__Examples:__

(1 + *i*)(2 + 2*i*) = 4*i*

(1 + *i*)(7 + 7*i*) = 14*i*

(1 + *i*)(0 + 0*i*) = 0*i*

[Remember that any number multiplied by zero(0) is still zero!]

(1 + *i*)(101 + 101*i*) = 202*i*

(1 + *i*)(√(2) + *i*√(2)) = 2*i*√(2)

(1 + *i*)(π + π*i*) = 2π*i*

## This math trick works with all numbers; however, if you pick an imaginary or complex number for *x*, then... Well, look at the examples below to find out:

(1 + *i*)(*i* + *i * i*) = (1 + *i*)(*i* + *i*^{2}) = (1 + *i*)(-1 + *i*) = -2

Let A = (1 + *i*); so, (1 + *i*)(A + A*i*) = (1 + *i*) * 2*i* = -2 + 2*i*

Let B = (1 - *i*); so, (1 + *i*)(B + B*i*) = (1 + *i*) * 2* *= 2 + 2*i*

Since *i*^{2} = -1, it affected the 2nd complex number in these 3 final examples. Also, the *real* part of a complex number is usually printed to the left of the plus or minus sign in-between. (A minus sign means that the *imaginary* part is negative; if you see a minus sign to the left of the *real* part, then the *real* part is negative.)

## However, the final examples are still correct according to the formula at the top of this Web page!

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© Derek Cumberbatch