CPhill

This is easier to show by video than by trying to describe it, but it's not difficult  !!!

https://www.youtube.com/watch?v=5bvjnleMn5A

Let a + bi  be the complex number.....then...

l a + bi  l    =   √ [ a² + b² ]

So

l 3 − 4i  l  =  √ [ 3² + (−4)² ]  = √ [ 25 ]   = 5

Let  p   be an odd = 2n + 1

So

p^2 + 1  =

[2n + 1] [2n + 1] + 1=

4n^2 + 4n + 1 + 1  =

4n^2 + 4n + 2  =

2 [2n^2 + 2n + 1 ]

Now let  [2n^2 + 2n + 1 ]  be an integer  = m

And  2m   is always even for any integer, m

Let n be he first integer and n + 1 the second.....so.....

n + [n + 1]  = 2n + 1   divide by 2

[ 2n + 1 ]  / 2 =    [  2n / 2  +  1/2 ]   = [  n + 1/2 ]

But....an even divided by 2 always produces an integer and no remainder.....thus.....2n + 1 must be odd

( n )  (n − 1)

If n is even, (n − 1) is odd.....and even * odd  = even

Else....if n is odd, (n − 1) is even.......and odd * even = even

2n + 1  ...... 2n is always even ......  and adding 1 to an even produces an odd.....

g(1/2)  = −(1/2)^2  + (1/2) + 3  =   −1/4 + 1/2 + 3  = 1/4 + 3  =3.25

[6 + 2/3 ]  − [3 + 1/2] =

[20/3] −  [7/2]  =

[ 40 − 21] / 6  =

[ 19 / 6] miles  =  3 + 1/6 miles   farther

It's pure luck.........

Answer here :  http://web2.0calc.com/questions/help-asap_24787