Question 688873
z*conj(z) = [modulus(z)]^2 


(a+bi)(a-bi) = [modulus(z)]^2 


a^2 - (bi)^2 = [modulus(z)]^2 


a^2 - b^2i^2 = [modulus(z)]^2 


a^2 - b^2(-1) = [modulus(z)]^2 


a^2 + b^2 = [modulus(z)]^2 


[sqrt(a^2 + b^2)]^2 = [modulus(z)]^2 


[modulus(z)]^2 = [modulus(z)]^2 


So z*conj(z) = [modulus(z)]^2  is true.


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z + conj(z) = 2a 


a+bi + a-bi = 2a 


(a+a)+(bi-bi) = 2a 


2a+0bi = 2a 


2a = 2a 


So z + conj(z) = 2a  is true.

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z - conj(z) = 2bi 


a+bi - (a-bi) = 2bi 


a+bi - a+bi = 2bi


(a-a)+(bi+bi) = 2bi


0a+2bi = 2bi


2bi = 2bi


So z - conj(z) = 2bi  is true.

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Therefore, the answer is choice A. All of the following.