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$\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!

M. Winter
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J. Doe
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  • Welcome to MSE. To type in MathJax, just use dollar signs around the script. – TPace Feb 15 '18 at 12:45
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    Let $z = x + iy$, continue from this. – IAmNoOne Feb 15 '18 at 12:45
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    No, you can't just assign random values, as that would just show that it is true for that particular pair of values. You have to prove it for all numbers. Try doing what Nameless suggested and see where you get stuck. – Bergson Feb 15 '18 at 12:54
  • I got 2x-2yi but I'm not sure where to go from there. – J. Doe Feb 15 '18 at 13:00

3 Answers3

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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.

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.

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.

Agile_Eagle
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  • I don't understand the last two dot points (I haven't done vectors yet), but I tried to use the first method below. Could you explain the methods please? – J. Doe Feb 15 '18 at 13:19
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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?

latbbltes
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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:

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

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

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

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

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

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

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