SOLUTION: Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbreviated conj(z) or conj(a + bi) is the complex number a-bi. Also, the modulus of

Question 688873: Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbreviated conj(z) or conj(a + bi) is the complex number a-bi. Also, the modulus of z, modulus(z) is the "size" of z, or SQRT(a2 + b2). Which of the following is true for all complex numbers?
A. All of the following
B. z*conj(z) = [modulus(z)]2
C. z + conj(z) = 2a
D. z - conj(z) = 2bi
E. None of the above
show your work please (:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
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.
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