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When trying to add $x$ to $x^{*}$ is it allowed to say that it would be equal to $2|x|$ i.e. so that $$x+x^{*}=2|x| $$ If this isn't the case is there any way to add them or should they be left as $$x+x^{*} $$

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If $x^*$ means the conjugate, more commonly writtin as $\bar x$, this is wrong. The right answer is twice the real part of $x$, written as $2\Re x$ or $2\operatorname{Re} x$ – Harald Hanche-Olsen Apr 7 '13 at 15:23
I've never seen anyone write the complex conjugate of $x$ as $x^*$ and I would recommend switching to the notation everyone else uses, unless perhaps you are taking a class and your prof uses this notation and expects you to use it. – Stefan Smith Apr 7 '13 at 16:38
up vote 4 down vote accepted

If we express $x\in \mathbb C$ and it's conjugate as follows:

$x = a + bi$, $\;\;x^* = \bar x = a - bi,\;\;$ where $\,a, b $ are real numbers.

Then adding $x$ to its conjugate gives us $$ x + \bar x \;=\; a + bi + a - bi\; =\; 2a, \;\;\; a \in \mathbb R$$

In other words, $\;x+ \bar x = 2\,\Re x,\;$ where $\;\Re x\;$ denotes the real part of the complex number $x$.

Note that $$2\sqrt{a^2 + b^2} \;= \;2|x| \;\;{\bf \neq}\;\;x + \bar x \;=\; 2a$$

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thx i see my mistake – user63407 Apr 7 '13 at 15:29
Can it make sense that $\Re(x)\pm\Re(y)=e^{\pm i T}$, it would imply that $\cos(T)=\Re(x)$ and $i\sin(T)=\Re(y)$ but I can't decide if it could be correct since one has a real term equalling another term multiplied with i? – user63407 Apr 7 '13 at 15:33
I don't think so, no. where does T come from, when equating $\Re(x) \pm \Re(y) = e^{\pm iT}$? $\Re(x)$ and $\Re(y)$ are simply real numbers. – amWhy Apr 7 '13 at 15:36
Thx its because I am trying to find the eigenvalues of $\lambda^{2}-A\lambda-A^{*}\lambda+1=0 $ and boil it down to $+\Re(A)\pm\sqrt{\Re(A)^{2}-1}$ and I am supposed to show that the answer is $=e^{\pm ix}$, where $\cos(x)=\Re(A)$. It is true that $|A|^{2}+|B|^{2}=1$ – user63407 Apr 7 '13 at 15:43
I spot my mistake the square root would be $\sqrt{-\Re(B)^{2}}$, – user63407 Apr 7 '13 at 15:45


$\bar a=(x-iy)$

$\bar a+a= 2x$.

$|a|$ is $\sqrt{x^2+y^2}$, therefore expression is wrong.

It has to be $\bar a+a=2\operatorname{Re} x$

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Let $a = (x+iy)$ where $x,y \in \mathbb{R}$ and $ i = \sqrt{-1}$. Then we separate this expression into real and imaginary parts respectively. The part without the imaginary unit $i$ is said to be the real component and the part with it is said to be imaginary. Also, there exists a unique conjugate $ \overline {a}$ such that $\overline {a} = x-iy$. When we add $a$ and its conjugate $\overline a$, notice that $a + \overline a = (x+iy) + (x-iy) \Rightarrow 2x$. Now notice how this expression only considers the real component.

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