If $a$ and $b$ are complex numbers then will the given condition stand TRUE? And if $a$ and $b$ are not complex numbers will the identity (a+b^2 be greater than or equal to zero?
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4The complex numbers are not an ordered field: there is no general notion of order between complex numbers. – Brian M. Scott Mar 10 '21 at 03:42
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1In $\mathbb{C}$ you can't even compare the order of terms so there is no $\ge 0$ – Aditya Dwivedi Mar 10 '21 at 03:43
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If you are referring to the magnitude of a complex number, then as long as $a$ and $b$ are not all equal to zero, then the magnitude of $(a + b)^2$ is greater than zero. – soupless Mar 10 '21 at 03:47
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1Beyond that, when allowing complex numbers it is possible for the square of the sum of two complex numbers to be a negative real number, for trivial example $(i+i)^2=-4$ – JMoravitz Mar 10 '21 at 03:47
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Nope. Obviously $(i+i)^2 = (2i)^2 = -4 < 0$. This is actually exactly why the complex numbers can not have order. (In an ordered field we have $a> 0 \implies a^2 > 0a =0$ and if $a< 0$ then $0 < -a$ so $a^2 =(-a)^2 > -a0 = 0$. and $0^2 =0$. So $a^2 \ge 0$ but $i^2 =-1$ and $1^2 =1$ so if $i^2=-1 > 0$ then $0 < 1$ but $1^2> 0$ too.... so no order is possible.) – fleablood Mar 10 '21 at 05:09
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What about $a=1+i$ and $b=-1+i$? There $(a+b)^2$ is negative and therefore less than 0.
The magnitude, however, of every complex number $c+di$; $c$, $d$ real, is $c^2+d^2$ which is always nonegstive real and positive if either $c,d$ is nonzero.
Mike
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