Proving $\bar z_1+\bar z_2=\overline{z_1+z_2}$ and other

Question

$\overline{z}_1+\overline{z}_2=\overline{z_1+z_2} $

Can I just assign two random values and prove it that way? (My textbook doesn't have an answer.)

I also have these problems, but I don't understand them either :(

$\frac{\overline{z}_1}{\overline{z}_2}=\overline{\left(\frac{z_1}{z_2}\right)}$

$\overline{z}_1\:\overline{z}_2=\overline{z_1z_2}$

If anyone can help, I would really appreciate it!

Answer 1

Hint for Method 1 (as told by Nameless) (algebraic)

Let $z = x + iy$

hence,

$\overline{z} = x - iy$

Now, try to work out each case.

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Hint for Method 2 (geometrical)

Use the following facts,

• Complex numbers can be represented on the Argand plane.
• Taking the modulus of a complex number is interpreted geometrically as taking its refection on the $X$ axis (real axis).
• Adding two complex numbers is just like adding two vectors.
• Multiplication of two complex numbers geometrically involves multiplying their magnitudes and adding their arguments.

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If you are still stuck or if you are only able to prove by one method, please comment and I will provide a detailed solution with both methods.

Answer 2

We can't just pick and random values as this statement is about every complex number, not just two of them.

So instead, lets right $z_1$ as $x_1+iy_1$ and $z_2$ as $x_2+iy_2$

Note that if $z=x+iy$ is a complex number, the definition of $\bar{z}$ is $x-iy$

So the left hand side can be written as $\bar{z_1}$+$\bar{z_2}$=$\overline{x_1+iy_1}+\overline{x_2+iy_2}$=$x_1-iy_1$+$x_2-iy_2$

Collecting like terms, we get that the LHS is equal to $x_1+x_2-iy_1-iy_2$

Now can you figure out how to simplify the RHS as well from this?

How can you show they are equal?

Answer 3

The problem you are talking is about the Well-Definition of an equality in maths.

At this link you can see what is the definition of "well defined" in mathematics.

First of all we have $\exists a,b,c,d\in\mathbb{R}$ such as $$\mathbb{C}\ni z_1=a+ib\implies\overline{z}_1=a-ib$$ $$\mathbb{C}\ni z_2=c+id \implies\overline{z}_2=c-id $$

Now let’s show that your formulas are well defined:

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$$\overline{z}_1+\overline{z}_2=\overline{z_1+z_2}\tag{1} $$

Then calculate the sum $$z_1+z_2=a+ib+c+id=(a+c)+i(b+d)\implies \\\overline{z_1+z_2}=\overline{(a+c)+i(b+d)}=(a+c)-i(b+d)=(a-ib)+(c-id)=\bar{z_1}+\bar{z_2}$$

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$$\overline{z}_1\overline{z}_2=\overline{z_1z_2}\tag{2} $$

Then calculate the product $$z_1z_2=(a+ib)(c+id)=ac+ida+ibc+i^2bd\overbrace{=}^{i^2=-1}(ac+bd)+i(ad+bc)\implies \\\overline{z_1z_2}=\overline{(ac+bd)+i(ad+bc)}=(ac+bd)-i(ad+bc)=(a-ib)(c-id)=\bar{z_1}\bar{z_2}$$

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$$\frac{\overline{z}_1}{\overline{z}_2}=\overline{\left(\frac{z_1}{z_2}\right)}\tag{3}$$

Then calculate the division: $$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{\frac{ac+bd}{c^2+d^2}+i \frac{cb -ad}{c^2+d^2}}=\frac{ac+bd}{c^2+d^2}-i \frac{cb -ad}{c^2+d^2}=\\=\frac{ac+(-b)(-d)}{c^2+(-d)^2}+i\frac{c(-b)-a(-d)}{c^2+(-d)^2}=\frac{a-ib}{c-id}=\frac{\bar{z_1}}{\bar{z_2}}$$ $$\tag*{$\square$}$$