Does this expression:

(C + C*)/ 2 = C


that C is always real?

Suppose C is complex: a+ib and thus C* = a-ib

Then we end up with

a = C and a is real, thus C is always real if it equals (C + C*)/ 2

  • $\begingroup$ Yes, your argument is fine. $\endgroup$ – angryavian Mar 10 '17 at 20:09
  • $\begingroup$ That's correct. Another way to write it is $c = \frac{1}{2}(c + c^*) = \Re(c) \in \mathbb{R}\,$. $\endgroup$ – dxiv Mar 10 '17 at 20:09
  • $\begingroup$ If C is mean to be a complex number (rather than $\mathbb C$ which is the set of all complex numbers) and $C^*$ is meant to be the complex conjugate (usually written as $\overline {C}$) then, yes, for any complex number $z$ then $z + \overline{z} = 2Re(z)$ is always a real number. By the way $z*\overline{z} = Re(z)^2 + Im(z)^2$ is also always a real number. $\endgroup$ – fleablood Mar 10 '17 at 20:11
  • $\begingroup$ Um @dxiv, Am I missing something $c = \frac 12(c + c^*)=\Re(c)$ isn't usually true, is it? Not if $c$ is complex? Did you mean to say $c = \frac 12(c+\overline{c}) + \frac 12(c - \overline{c})=\Re(c) + i\Im(c)$? $\endgroup$ – fleablood Mar 10 '17 at 20:15
  • $\begingroup$ @fleablood $c=\frac{1}{2}(c+c^*)$ is the premise of the exercise. $\endgroup$ – dxiv Mar 10 '17 at 20:18

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