Hi elsmith,
w is the complex conjugate of z (and vice versa), and we are asked to prove that w^2 is equal to the complex conjugate of z^2.
If z = a + bi and w = c + di, the complex conjugate of z is a - bi (that's what complex conjugate is, same real part, the negative of the imaginary part).
Then c = a and d = -b. w = a + -bi.
Square z. cc(z) I define to mean the complex conjugate of z. cc(z^2) = cc(a + bi)^2
= cc(a + bi)(a + bi) Now, use FOIL.
= cc(a^2 + abi + abi + (b^2)(i^2)) I squared is -1, so the last term is the same as -b^2.
= cc(a^2 - b^2 + 2abi)
= a^2 - b^2 - 2abi
Now, square w. w^2 = (a + -bi)^2
= (a + -bi)(a + -bi)
= a^2 - abi - abi + ((-b)^2)(i^2)
= a^2 - b^2 - 2abi.
So, z^2 and w^2 are indeed equal, proving the claim.
If you have any further questions, feel free to message me!
Sincerely, taeuknam.
