MHB Is z + z¯ and z × z¯ Real for Any Complex Number z?

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For any complex number z, both z + z¯ and z × z¯ are confirmed to be real. The expression z + z¯ simplifies to 2a, where a is the real part of z. The product z × z¯ results in a^2 + b^2, which is also a real number since both a and b are real numbers. The solution provided is accurate and demonstrates that these operations yield real results for any complex number. Thus, the claim is validated.
Yordana
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I apologize in advance for my English.
I want to know if my solution is correct. :)

To verify that for every complex number z, the numbers z + z¯ and z × z¯ are real.

My solution:
z = a + bi
z¯ = a - bi
z + z¯ = a + bi + a - bi = 2a ∈ R
z × z¯ = (a + bi) × (a - bi) = a^2 + b^2 ∈ R
 
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Yep. All correct.
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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