Recall that the symbol z represents the complex conjugate of z. if z = a + bi and w = c + di, prove the statement. z2 = z2 working from the left-hand side, simplify the expression into the standard complex form.

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.

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.