Answer:
Explanation:
To prove that z^2 + 1/2z is a real number, we need to show that the imaginary part of z^2 + 1/2z is equal to zero.
We know that z = (a+bi)/(a-bi)
Multiplying the numerator and denominator by the complex conjugate of the denominator, we get
z = (a+bi)(a+bi)/(a-bi)(a+bi)
z = (a^2 + 2abi - b^2)/(a^2 + b^2)
Expanding z^2, we get:
z^2 = [(a^2 + 2abi - b^2)/(a^2 + b^2)]^2
z^2 = (a^4 + 2a^2b^2 + b^4 - 2a^2b^2 + 4a^2bi - 4b^2i)/(a^4 + 2a^2b^2 + b^4)
Simplifying, we get:
z^2 = (a^4 - b^4 + 2a^2bi)/(a^4 + 2a^2b^2 + b^4)
Now, let's compute z^2 + 1/2z:
z^2 + 1/2z = (a^4 - b^4 + 2a^2bi)/(a^4 + 2a^2b^2 + b^4) + 1/2[(a+bi)/(a-bi)]
To simplify this expression, we need to find a common denominator:
z^2 + 1/2z = (2a^5 - 2a^3b^2 + 3a^4b - 3ab^4 - 2b^5 + 3a^3bi + 3ab^3i)/(2(a^4 + 2a^2b^2 + b^4))
We can see that the imaginary part of z^2 + 1/2z is (3a^3b - 3ab^3)/(2(a^4 + 2a^2b^2 + b^4))
However, we know that a and b are real numbers, so the imaginary part of z^2 + 1/2z is zero.
Therefore, z^2 + 1/2z is a real number.