[You forgot to tell us what f(x) is. So I made up one that
would work the same way, f(x) = x²+4x+1.
To show that f(-2+2i) is a real number, we substitute (-2+2i)
for x in
f(x) = x²+4x+1
f(-2+2i) = (-2+2i)²+4(-2+2i)+1
f(-2+2i) = (-2+2i)(-2+2i)-8+8i+1
f(-2+2i) = (4-4i-4i+4i²)-8+8i+1
f(-2+2i) = 4-4i-4i+4i²-8+8i+1
f(-2+2i) = 4-8i+4i²-8+8i+1
Since i² = -1, we can substitute (-1) for i²
f(-2+2i) = 4-8i+4(-1)-8+8i+1
f(-2+2i) = 4-8i-4-8+8i+1
The -8i and the 8i cancel out.
f(-2+2i) = 4-4-8+1
f(-2+2i) = -7
And -7 is a real number because it does not have any i's,
since the terms that had i's all canceled out.
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To show that f(7+4i) is NOT a real number, we substitute (7+4i)
for x in
f(x) = x²+4x+1
f(7+4i) = (7+4i)²+4(7+4i)+1
f(7+4i) = (7+4i)(7+4i)+28+16i+1
f(7+4i) = (49+28i+28i+16i²)+28+16i+1
f(7+4i) = 49+28i+28i+16i²+28+16i+1
f(7+4i) = 49+56i+16i²+28+16i+1
Since i² = -1, we can substitute (-1) for i²
f(7+4i) = 49+56i+16(-1)+28+16i+1
f(7+4i) = 49+56i-16+28+16i+1
f(7+4i) = 49-16+28+1+56i+16i
f(7+4i) = 62+72i
There is still an i in 62+72i, so that is NOT a real number.
Now do the same thing with the equation for f(x) which you
were given that you forgot to type above.
Edwin
