Why is the product of a complex number and its conjugate always a real number?

Answer 1

because (a + bi)*(a - bi) = a^2 -abi + abi -i^2b^2 = a^2 + b^2

Here, a and b are real, so a^2 + b^2 is not only real, but nonnegative.

Answer 2

The simple answer is that when they are multiplied the imaginary parts cancel out to zero leaving only the real part.
Example (3 +5i)(3 - 5i) = 9 + 15i - 15i -25i^2 = 9 + 25 = 34

Answer 3

Let a + bi be the complex number. Its conjugate is a - bi. The product is as follows:

(a+bi)(a-bi) = a² - abi + bai - b²(i)²

The terms - abi and + bai cancel out, and (i)² becomes -1, so we're left with

a² + b², always a positive number.

Answer 4

The important property of complex numbers
is that i² = - 1.
Consider the product :-
(a + i b).(a - i b)
= a² - iab + iba - i²b²
= a² - iab + iab - (-1)b²
= a² + b² (a real number)

Answer 5

If z= a + ib, then the conjugate is a - ib, so

(a+ib)(a-ib) = a^2-(ib)^2 = a^2 - (-1)b^2 = a^2 + b^2

Answer 6

see for yourself:

(x + a)(x - a) = x² + ax - ax - a² = x² - a²
the inner and outer products add to 0

(x + ai)(x - ai) = x² + axi - axi - a²i² = x² - a²i²

but i² = -1, so x² - a²i² = x² -a²(-1) = x² + a²

Answer 7

because (a+bi)(a-bi)=a^2+b^2; a,b R