Prove that $\overline{a+b}=\bar{a}+\bar{b}$

Question

Question:

Prove that $\overline{a+b}=\bar{a}+\bar{b}$

My attempt: I've tried this:

$\overline{a+b}=\bar{a}+\bar{b}$ $$ (x+iy)-(z+in)=(x-iy)+(z-in)= (x-z)+(y-n)=(x+z)+(-y-n) $$ But I just can't get the right answer, I would be glad if you could explain this to me.

Answer 1

$a = x + i y$ and $b = u + iv$, then \begin{align} \overline{a + b} & = \overline{x + i y + u + i v} \\ & = \overline{x + u + i(y + v)} \\ & = x + u - i(y+v) \\ & = x - iy + u - iv \\ & = \bar a + \bar b \end{align}

Answer 2

You seem to have defined $a=x+iy, b=z+in$. Then $\overline {a+b} = \overline {x+iy+z+in}=x-iy+z-in=\overline a + \overline b$

Answer 3

Let $a=x_1+y_1i$ and $b=x_2+y_2i$, where $\{x_1,x_2,y_1,y_2\}\subset\mathbb R$.

Thus, $$\overline{a+b}=\overline{ x_1+x_2+(y_1+y_2)i}= x_1+x_2-(y_1+y_2)i=x_1-y_1i+x_2-y_2i=\overline{a}+\overline{b}.$$ Done!