Linear Algebra, Triangle Inequality Justification

Question

So I get that the triangle inequality shows that $|z_1 + z_2| \leq |z_1| + |z_2|$.But I do not understand how,

$$z_2 \overline{z_1} + z_1 \overline{z_2} = 2\Re(z_2 \overline{z_1}) \leq 2∣z_2 \overline{z_1}∣ = 2∣z_2∣ \cdot ∣\overline{z_1}∣ = 2∣z_2∣ \cdot ∣z_1∣$$

More specifically, the beginning part of how $z_2 \overline{z_1} + z_1 \overline{z_2} = 2\Re(z_2 \overline{z_1})$. Could someone explain this please?

Answer 1

Let $z = x + yi\in\mathbb{C}$. Then we get $z + \overline{z} = (x + yi) + (x - yi) = 2x = 2\text{Re}(z)$.

We do also have that $|\overline{z}| = \sqrt{x^{2} + (-y)^{2}} = \sqrt{x^{2} + y^{2}} = |z|$.

Finally, the following result holds:

\begin{align*} \text{Re}(z) = x \leq |x| = \sqrt{x^{2}} \leq \sqrt{x^{2} + y^{2}} = |z| \end{align*}

Gathering all these claims, one gets the desired relation.

Hopefully this helps!

Answer 2

For the future, you should pick up some latex so that you can ask questions more easily. It's always nicer for someone to respond when they can read what you've written.

Let $z = x+iy$ be a complex number. Then, $x = \Re(z)$. So: $$|\Re(z)| \leq \sqrt{|x|^2} \leq \sqrt{|x|^2+|y|^2} = |z|$$ Next, $\overline{z} = x-iy$. So: $$z +\overline{z} = 2x = 2 \Re(z)$$

If we have two complex numbers $z$ and $w$, then we know that: $$\overline{zw} = \overline{z} \cdot \overline{w}$$ In your case, this implies that: $$\overline{z_2} z_1 = \overline{z_2 \overline{z_1}}$$ $$z_1 \overline{z_2} + \overline{z_1} z_2 = \overline{\overline{z_1}z_2} + \overline{z_1}z_2 = 2\Re(\overline{z_1}z_2)$$ as was desired.

Answer 3

Let $z_1=x_1+iy_1$, $z_2=x_2+iy_2$ then $$ \begin{align} z_1 \overline{z_2} &= (x_1+iy_1)(x_2-iy_2) \\ &=x_1 x_2 + i(x_2y_1 - x_1 y_2)+y_1y_2\\ \overline{z_1}z_2 &=(x_1-iy_1)(x_2+iy_2)\\ &=x_1 x_2 + i(x_1 y_2-x_2y_1)+y_1y_2\\ z_1 \overline{z_2} + \overline{z_1}z_2 &=2x_1x_2+2y_1y_2\\ &=2\Re(z_1\overline{z_2}) \end{align} $$

Answer 4

It comes from the "twisted nature" of multiplying complex numbers.

If you multiply two complex numbers the Real part is the product of the real parts minus the product of the imaginary parts

That is: $Re(zw) = [Re(z)Re(w) - Im(z)Im(w)]$

and the imaginary part is the sum of the product of one imaginary part and the other real part and the product one real part and the other imaginary part

That is: $Im(zw) = [Im(z)Re(w) +Re(z)Im(w)]$

Proof: Just do it.

If $z = a+bi$ and $w=\alpha + \beta i$ then

$(a+bi)(\alpha + \beta i) = a\alpha + bi\cdot \alpha +a\cdot \beta i + bi\cdot \beta i=$
$a\alpha + \alpha bi + a\beta i + b\beta i^2=$

$i^2 = -1$ so $b\beta i^2$ becomes $-b\beta$ and that's now a real part so we put it with the other product of reals to get $a\alpha -b\beta$. That is how we can get that "difference of the products of the extremes".

Meanwhile the $i$s distribute and products of a real with an imaginary gives as a component of the imaginary so $\alpha bi + a\beta i = (\alpha b + a\beta)i$ and that is how we can get that "sum of the products of the middles".

So $a\alpha + \alpha bi + a\beta i + b\beta i^2=(a\alpha -b\beta) +(a\beta + \alpha b)i$.

And $Re(zw) =(a\alpha -b\beta)= [Re(z)Re(w) - Im(z)Im(w)]$ and

$Im(zw) = (a\beta + \alpha b) = [Re(z)Re(w) - Im(z)Im(w)]$

......

Now the whole idea of the Conjugate is that you flip the sign of the imaginary part.

So if whereas $Re(zw) = Re(z)Re(w) - Im(z)Im(w)$ we'd have, when we calculate $z\overline w$ that $Im(\overline w) = -Im(w)$ but also $Re(\overline w)=Re(\overline w)$ and we get

$Re(z\overline w) = Re(z)Re(w)- Im(z)Im(\overline w)=$
$Re(z)Re(w) - Im(z)\cdot[-Im(w)]=$ $Re(z)Re(w) + Im(z)Im(w)$.

And we get that $Im(z\overline w) = Im(z)Re(\overline w) + Im(z)Im(\overline w)=$
$Im(z)Re(w) +Im(z)[-Im(w)] = Im(z)Re(w)-Im(z)Im(w)$.

So

$Re(z\overline w) =Re(z)Re(w) + Im(z)Im(w)$ while $Im(z\overline w)=Im(z)Re(w) - Re(z)-Im(z)Re(w)$.

As addition is commutative but subtraction isn't (reversing the order of subtract results in the opposite sign!) we get

$Re(w\overline z) = Re(w)Re(z) + Im(w)Im(z)=Re(z\overline w)$ but $Im(w\overline z) = Im(w)Re(z) - Re(w)Im(z) = -[Im(z)Re(w) - Re(z)Im(w)] = -Im(z\overline w)$.

And there for if we add $w\overline z + z\overline w$ the two real parts (which are eqaul) would compound each other with the two imaginary parts (which are opposites) will cancel each other out.

So $z\overline w + w\overline z = $
$[(Re(z)Re(w) + Im(z)Im(w)) + (Im(z)Re(w)-Re(z)Im(w))i] +$
$[(Re(w)Re(z) + Im(w)Im(z)) + (Im(w)Re(z)-Re(w)Im(z))i]= $

$\{(Re(z)Re(w) + Im(z)Im(w))+(Re(w)Re(z) + Im(w)Im(z))\} +\{(Im(z)Re(w)-Re(z)Im(w))+(Im(w)Re(z)-Re(w)Im(z))\}i = $
$2(Re(z)Re(w) + Im(z)Im(w))+ 0\cdot i =$
$2Re(z)Re(w) = Re(z\overline w)= Re(w\overline z)$.

.....

But don't just take me word for it.

Do it!

If $z = a+ bi$ and $w = \alpha + \beta i$ then what IS $z\overline w + w\overline z$?

Answer:

$z\overline w= (a+bi)(\alpha -\beta i) = (a\alpha + b\beta) + (b\alpha -a\beta)i$

and

$w\overline z =(\alpha + \beta i)(a+bi) = (a\alpha + b\beta) + (a\beta -b\alpha)i$

and so $z\overline w+w\overline z=$

$[(a\alpha + b\beta)+(a\alpha + b\beta)] + [(b\alpha -a\beta)+ (a\beta -b\alpha)]i=$
$2(a\alpha + b\beta) + 0\cdot i = 2(a\alpha +b\beta) =2Re(z\overline w)=2Re(w\overline z)$.