# Math for Non-Geeks: Absolute value and conjugation

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## Absolute value of a complex number
### Motivation of the absolute value

When dealing with real numbers, we introduced the absolute value function $|\cdot |\colon \mathbb {R} \to \mathbb {R} _{\geq 0}$, with which we could specify the distance of a certain number to zero. Visualised on the real number line this looks as follows:
Also in the complex plane we can determine the distance of a complex number from the zero point. For this we use the theorem of Pythagoras. Let $z=a+b\,\mathrm {i}$ be a complex number:
With the Pythagorean theorem, for the distance $|z|$ from zero, we have $|z|^{2}=\operatorname {Re} (z)^{2}+\operatorname {Im} (z)^{2}=a^{2}+b^{2}$. By taking roots on both sides, $|z|$ can be determined. There is:
$|z|={\sqrt {{\text{Re}}(z)^{2}+{\text{Im}}(z)^{2}}}={\sqrt {a^{2}+b^{2}}}$
The absolute value transfers some concepts of real numbers to the complex numbers. As $|x-y|$ in the real numbers is the distance between $x$ and $y$, so is $|z-w|$ in the complex numbers the distance between $z$ and $w$. The distance in turn can be used to define terms such as a limit: A complex number $z$ is the limit value of a sequence $(z_{n})_{n\in \mathbb {N} }$ of complex numbers, if the distance $|z-z_{n}|$ between the limit value $z$ and the sequence elements $z_{n}$ becomes arbitrarily small (below any $\epsilon$).

### Definition of the complex absolute value

Definition (absolute value of a complex number)
Let $z=a+b\,\mathrm {i} \in \mathbb {C}$. Then, we define $|z|={\sqrt {{\text{Re}}(z)^{2}+{\text{Im}}(z)^{2}}}={\sqrt {a^{2}+b^{2}}}$ and call the number $|z|\in \mathbb {R} _{\geq 0}$ the absolute value of $z$.
Hint
The absolute value defined above on the complex numbers coincides to the usual absolute value for real numbers. Consider $z=a+0\cdot \mathrm {i} \in \mathbb {R}$. Then, there is:
${\begin{aligned}|z|&={\sqrt {{\text{Re}}(z)^{2}+{\text{Im}}(z)^{2}}}={\sqrt {a^{2}+0^{2}}}={\sqrt {a^{2}}}\\[0.3em]&={\begin{cases}a,&{\text{if }}a\geq 0,\\-a,&{\text{if }}a<0.\end{cases}}\end{aligned}}$

## Complex conjugation
### Motivation of the conjugation

The imaginary unit $\mathrm {i}$ as a root of $-1$ fulfils the equation $\mathrm {i} ^{2}=-1$. We can imagine the multiplication $x\mapsto -1\cdot x$ as a $180^{\circ }$ rotation around the zero point. Now because of the equation $\mathrm {i} ^{2}=-1$, the multiplication $x\mapsto -1\cdot x$ is the same as $x\mapsto \mathrm {i} \cdot (\mathrm {i} \cdot x)$. Thus, $x\mapsto \mathrm {i} \cdot x$ is an operation that corresponds to a $180^{\circ }$ rotation when used twice.
If a double multiplication with $\mathrm {i}$ corresponds to a $180^{\circ }$ rotation, then a single multiplication must correspond to a $90^{\circ }$ rotation. In particular, the imaginary unit $\mathrm {i}$ is equal to the number which results from a rotation of the number $1$ by $90^{\circ }$:
It is common in Mathematics to turn counter-clockwise. Thus, $\mathrm {i}$ is located where in $\mathbb {R} \times \mathbb {R}$ the $1$ lies on the $y$ axis. However, one could just as well have turned clockwise. Then the $\mathrm {i}$ would lie at the position of the $-1$ on the $x$ axis:
We could also have derived the complex numbers from this alternative rotation. In that case we would have obtained a different set of complex numbers where the imaginary unit $\mathrm {i}$ is below the $x$ axis. In this alternative set of complex numbers the roles of $\mathrm {i}$ and $-\mathrm {i}$ are reversed. So if we swap $\mathrm {i} \leftrightarrow -\mathrm {i}$ everywhere, essential properties and structures obtained by number range expansion should be preserved. Such an interchange is shown in the figure:
$a+b\,\mathrm {i} \mapsto a-b\mathrm {i}$
In this figure, the imaginary part is multiplied by $-1$. This corresponds to a reflection of the complex number on the real ($x$) axis:
An example where this reflection is useful are the zeros of the function $f\colon \mathbb {C} \to \mathbb {C} ,f(z)=z^{2}+1$. There is $f(\mathrm {i} )=\mathrm {i} ^{2}+1=0$. Therefore $\mathrm {i}$ is a zero of $f$. On the other hand there is also $f(-\mathrm {i} )=(-\mathrm {i} )^{2}+1=0$ and therefore $-\mathrm {i}$ is another zero. Let us consider $g(z)=z^{2}-2z+2$ with the zero $1+\mathrm {i}$. One might think that the negative of the number, $-1-\mathrm {i}$, is another zero. Unfortunately this is not the case. But if we exchange $\mathrm {i}$ with $\mathrm {-} i$, i.e. if we look at the complex number $1-\mathrm {i}$, we get another zero:
${\begin{aligned}g(1-\mathrm {i} )&=(1-\mathrm {i} )^{2}-2(1-\mathrm {i} )+2\\&=(-2\mathrm {i} )-(2-2\mathrm {i} )+2=0\end{aligned}}$
For die Nullstelle $z=a+b\,\mathrm {i}$ eines Polynoms scheint das an der reellen Achse gespiegelte $a-b\,\mathrm {i}$ eine weitere Nullstelle zu sein. Dies ist im Übrigen for all Polynome with rein reellen Koeffizienten der Fall. Dies weist darauf hin, dass die Abbildung $a+b\,\mathrm {i} \mapsto a-b\,\mathrm {i}$ eine Besondere ist. Diese Abbildung wird complex Konjugation genannt.
For the zero point $z=a+b\,\mathrm {i}$ of a polynomial, the $a-b\,\mathrm {i}$ mirrored on the real axis seems to be another zero point. This is the case for all polynomials with purely real coefficients. This indicates that the mapping $a+b\,\mathrm {i} \mapsto a-b\,\mathrm {i}$ might be particularly useful. As you progress with your math studies, you will see that it is useful in many more situations. So we better give it a name. Let's call it complex conjugation (since it connects/ conjugates $a+b\,\mathrm {i}$ and $a-b\,\mathrm {i}$).

### Definition of the complex conjugation

Definition (complex conjugation for a complex number)
Let $z=a+b\,\mathrm {i} \in \mathbb {C}$. Then, the mapping $\mathbb {C} \to \mathbb {C} ,z\mapsto {\bar {z}}$ is called complex conjugation and the number ${\bar {z}}=a-b\,\mathrm {i}$ is the complex conjugate of $z$.

## Overview: Properties of the absolute value and the complex conjugation
### Properties of the complex conjugation

For all $z$ and $w\in \mathbb {C}$ there is:
* $z={\overline {z}}\iff \operatorname {Im} (z)=0\iff z\in \mathbb {R}$
* ${\overline {\overline {w}}}=w$
* ${\overline {z+w}}={\overline {z}}+{\overline {w}}$
* ${\overline {z\cdot w}}={\overline {z}}\cdot {\overline {w}}$
* ${\overline {\sum _{k=1}^{n}z_{k}}}=\sum _{k=1}^{n}{\overline {z_{k}}}$
* ${\overline {\prod _{k=1}^{n}z_{k}}}=\prod _{k=1}^{n}{\overline {z_{k}}}$
* ${\text{Re}}(w)={\frac {1}{2}}\cdot \left(w+{\overline {w}}\right)$
* ${\text{Im}}(w)={\frac {1}{2\mathrm {i} }}\cdot \left(w-{\overline {w}}\right)$
* $|z|={\sqrt {z\cdot {\bar {z}}}}$
* $z^{-1}={\frac {\overline {z}}{|z|^{2}}}$
* ${\overline {\left({\frac {z}{w}}\right)}}={\frac {\overline {z}}{\overline {w}}}$

### Properties of the absolute value of a complex number

For all $z$ and $w\in \mathbb {C}$ there is:
* $|z|\geq 0$ and $|z|=0\Leftrightarrow z=0$ (positive definiteness)
* $|wz|=|w||z|$ (multiplicativity)
* $|{\text{Re}}(z)|\leq |z|$ and $|{\text{Im}}(z)|\leq |z|$
* $|w+z|\leq |w|+|z|$ (triangle inequality)
* $|z|\leq |{\text{Re}}(z)|+|{\text{Im}}(z)|$
* ${\big |}|w|-|z|{\big |}\leq |w-z|$

## Computation rules for complex conjugation
### Conjugation does not change real numbers

Theorem (Conjugation does not change real numbers)
For a number $z\in \mathbb {C}$ there is ${\bar {z}}=z$ if and only if $z$ is purely real, i.e. $\operatorname {Im} (z)=0$.
Proof (Conjugation does not change real numbers)
Proof step: $z={\overline {z}}\implies z\in \mathbb {R}$
Let $z=a+b\,\mathrm {i}$ with $a,b\in \mathbb {R}$ and ${\overline {z}}=z$. There is then
${\begin{array}{rrl}&z&{\overline {z}}\\\implies &a+b\,\mathrm {i} &=a-b\,\mathrm {i} \\\implies &b\,\mathrm {i} &=-b\,\mathrm {i} \\\implies &2b\,\mathrm {i} &=0\\\implies &b&=0\end{array}}$
Hence $\operatorname {Im} (z)=b=0$ so $z$ is a real number.
Proof step: $z\in \mathbb {R} \implies z={\overline {z}}$
Let $z=a+b\,\mathrm {i}$ be a real number. That means, $\operatorname {Im} (z)=b=0$. We have:
${\overline {z}}=a-b\,\mathrm {i} =a-0\mathrm {i} =a+0\mathrm {i} =z$

### Involution

Theorem (Involution)
For a complex number $w\in \mathbb {C}$ there is:
${\overline {\overline {w}}}=w$
Proof (Involution)
Let $w=a+b\,\mathrm {i}$ with $a,b\in \mathbb {R}$. Then, there is:
${\overline {\overline {w}}}={\overline {(a-b\,\mathrm {i} )}}=a+b\,\mathrm {i} =w$
This can also be explained as follows: ${\overline {w}}$ is the reflection of $w$ on the real axis. So ${\overline {\overline {\overline {w}}}}$ is the reflection of the reflection, which is just the original complex number.

### Compatibility with addition

Theorem (Compatibility with addition)
For complex numbers $z,w\in \mathbb {C}$ there is:
${\overline {z+w}}={\overline {z}}+{\overline {w}}$
Proof (Compatibility with addition)
Let $z\in \mathbb {C}$ be of the form $a+b\,\mathrm {i}$, where $a,b\in \mathbb {R}$ and $w\in \mathbb {C}$ of the form $c+d\,\mathrm {i}$, where $c,d\in \mathbb {R}$. Then, there is:
${\begin{aligned}{\overline {z+w}}&={\overline {(a+b\,\mathrm {i} )+(c+d\,\mathrm {i} )}}\\&={\overline {a+b\,\mathrm {i} +c+d\,\mathrm {i} }}\\&={\overline {(a+c)+(b+d)\,\mathrm {i} }}\\&=(a+c)-(b+d)\,\mathrm {i} \\&=(a+c)-b\,\mathrm {i} -d\,\mathrm {i} \\&=a-b\,\mathrm {i} +c-d\,\mathrm {i} \\&=(a-b\,\mathrm {i} )+(c-d\,\mathrm {i} )\\&={\overline {z}}+{\overline {w}}\end{aligned}}$

### Compatibility with multiplication

Theorem (Compatibility with multiplication)
For complex numbers $z,w\in \mathbb {C}$ there is:
${\overline {z\cdot w}}={\overline {z}}\cdot {\overline {w}}$
Proof (Compatibility with multiplication)
Let $z\in \mathbb {C}$ be of the form $a+b\,\mathrm {i}$, where $a,b1\in \mathbb {R}$ and $w\in \mathbb {C}$ of the form $c+d\,\mathrm {i}$, where $c,d\in \mathbb {R}$. Then, there is:
${\begin{aligned}{\overline {z\cdot w}}&={\overline {(a+b\,\mathrm {i} )\cdot (c+d\,\mathrm {i} )}}\\&={\overline {ac+ad\,\mathrm {i} +bc\,\mathrm {i} -bd}}\\&={\overline {(ac-bd)+(ad+bc)\,\mathrm {i} }}\\&=ac-bd-(ad+bc)\,\mathrm {i} \\&=ac-bd-ad\,\mathrm {i} -bc\,\mathrm {i} \\&=ac-ad\,\mathrm {i} -bc\,\mathrm {i} -bd\\&=(a-b\,\mathrm {i} )\cdot (c-d\,\mathrm {i} )\\&={\overline {z}}\cdot {\overline {w}}\end{aligned}}$

### Compatibility of conjugation with finite sums and products

We know how the conjugation behaves with the sum and product of two numbers. What happens with sums and products with three or more numbers like ${\overline {z_{1}+z_{2}+z_{3}}}$? We use a trick: First, we consider $(z_{1}+z_{2})$ as a single complex number. Then,we twice use the theorem of conjugation and sum:
${\overline {(z_{1}+z_{2})+z_{3}}}={\overline {(z_{1}+z_{2})}}+{\overline {z_{3}}}={\overline {z_{1}}}+{\overline {z_{2}}}+{\overline {z_{3}}}$
There is also no difference for three summands if we first sum everything and then apply conjugation to the resulting number, or if we first conjugate every number and then sum everything. This is generally true for arbitrary sums and products of complex numbers, as we will prove below via induction.
Theorem (Compatibility for arbitrarily many complex numbers)
For every $n\in \mathbb {N}$ and all complex numbers $z_{1},...z_{n}\in \mathbb {C}$ there is:
* ${\overline {\sum _{k=1}^{n}z_{k}}}=\sum _{k=1}^{n}{\overline {z_{k}}}$
* ${\overline {\prod _{k=1}^{n}z_{k}}}=\prod _{k=1}^{n}{\overline {z_{k}}}$
Proof (Compatibility for arbitrarily many complex numbers)
We prove this theorem for the sum via full induction. The proof for the finite product can be done analogously.
Theorem whose validity shall be proven for $n\in \mathbb {N}$ bewiesen werden soll:
${\overline {\sum _{k=1}^{n}z_{k}}}=\sum _{k=1}^{n}{\overline {z_{k}}}$
1. Base case:
${\overline {\sum _{k=1}^{1}z_{k}}}={\overline {z_{1}}}=\sum _{k=1}^{1}{\overline {z_{k}}}$
2. Inductive step:
2a. Inductive hypothesis:
${\overline {\sum _{k=1}^{n}z_{k}}}=\sum _{k=1}^{n}{\overline {z_{k}}}$
2b. Induction theorem:
${\overline {\sum _{k=1}^{n+1}z_{k}}}=\sum _{k=1}^{n+1}{\overline {z_{k}}}$
2c. Proof of induction step:
${\begin{aligned}{\overline {\sum _{k=1}^{n+1}z_{k}}}&={\overline {\sum _{k=1}^{n}z_{k}+z_{n+1}}}={\color {OliveGreen}{\overline {\sum _{k=1}^{n}z_{k}}}}+{\overline {z_{n+1}}}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{induction assumption}}\right.}\\[0.5em]&={\color {OliveGreen}\sum _{k=1}^{n}{\overline {z_{k}}}}+{\overline {z_{n+1}}}=\sum _{k=1}^{n+1}{\overline {z_{k}}}\end{aligned}}$

### Computation of the real and imaginary part

Theorem (Computation of the real and imaginary part)
For a complex number $w\in \mathbb {C}$ there is:
* ${\text{Re}}(w)={\frac {1}{2}}\cdot \left(w+{\overline {w}}\right)$
* ${\text{Im}}(w)={\frac {1}{2\mathrm {i} }}\cdot \left(w-{\overline {w}}\right)$
Proof (Computation of the real and imaginary part)
Let $w=a+b\,\mathrm {i}$ with $a,b\in \mathbb {R}$. We verify this equation by direct computation, starting from the right:
${\begin{aligned}{\frac {1}{2}}\cdot \left(w+{\overline {w}}\right)&={\frac {1}{2}}\cdot (a+b\,\mathrm {i} +(a-b\,\mathrm {i} ))\\[0.3em]&={\frac {1}{2}}\cdot (a+a+b\,\mathrm {i} -b\,\mathrm {i} )\\[0.3em]&={\frac {1}{2}}\cdot (2a)=a=\mathrm {Re} (w)\\[0.3em]{\frac {1}{2\mathrm {i} }}\cdot \left(w-{\overline {w}}\right)&={\frac {1}{2\mathrm {i} }}\cdot (a+b\,\mathrm {i} -(a-b\,\mathrm {i} ))\\[0.3em]&={\frac {1}{2\mathrm {i} }}\cdot (a-a+b\,\mathrm {i} +b\,\mathrm {i} )\\[0.3em]&={\frac {1}{2\mathrm {i} }}\cdot (2b\,\mathrm {i} )=b=\mathrm {Im} (w)\end{aligned}}$

### Computation of the absolute value via conjugation

Theorem (Computation of the absolute value via conjugation)
For all complex numbers $z\in \mathbb {C}$ there is $|z|={\sqrt {z\cdot {\bar {z}}}}$. So $z\cdot {\overline {z}}=|z|^{2}$.
Proof (Computation of the absolute value via conjugation)
Let $z\in \mathbb {C}$ be an arbitrary complex number with $z=a+b\,\mathrm {i}$. We compute $z\cdot {\bar {z}}$. There is $z=a+b\,\mathrm {i}$ and ${\bar {z}}=a-b\,\mathrm {i}$. So there is
${\begin{aligned}z\cdot {\bar {z}}&=(a+b\,\mathrm {i} )\cdot (a-b\,\mathrm {i} )=a^{2}-(b\,\mathrm {i} )^{2}\\&=a^{2}-(-b^{2})=a^{2}+b^{2}=|z|^{2}\end{aligned}}$
Since $|z|^{2}$ is real and the basis $|z|$ is non-negative,m we can take the root and obtain the real number $|z|={\sqrt {z\cdot {\bar {z}}}}$.

### Compoutation of the inverse via conjugation

Theorem (Compoutation of the inverse via conjugation)
For all complex numbers $z\in \mathbb {C} \setminus \{0\}$ there is $z^{-1}={\tfrac {\overline {z}}{|z|^{2}}}$.
Proof (Compoutation of the inverse via conjugation)
Let $z\in \mathbb {C} \setminus \{0\}$ be an arbitrary complex number. We want to show $z^{-1}={\tfrac {\overline {z}}{|z|^{2}}}$. This is done by proving $z\cdot {\tfrac {\overline {z}}{|z|^{2}}}=1$. With $|z|^{2}=z\cdot {\bar {z}}$ we get
$z\cdot {\frac {\overline {z}}{|z|^{2}}}={\frac {z\cdot {\overline {z}}}{|z|^{2}}}={\frac {|z|^{2}}{|z|^{2}}}=1=z\cdot z^{-1}$
Because the inverse is unique in a field, $z^{-1}={\tfrac {\overline {z}}{|z|^{2}}}$ follows. This proves that ${\tfrac {\overline {z}}{|z|^{2}}}$ is equal to the inverse of $z$.
Hint
When proving that the complex numbers form a field, we have also derived a multiplicative inverse of $z=a+b\,\mathrm {i}$. There we have seen $z^{-1}={\frac {a}{a^{2}+b^{2}}}-{\frac {b}{a^{2}+b^{2}}}\,\mathrm {i}$. This is consistent with the new representation by complex conjugation:
$z^{-1}={\frac {a}{a^{2}+b^{2}}}-{\frac {b}{a^{2}+b^{2}}}\,\mathrm {i} ={\frac {a-b\,\mathrm {i} }{a^{2}+b^{2}}}={\frac {\overline {z}}{|z|^{2}}}.$

### Conjugation of fractions

Theorem (Conjugation of fractions)
For all complex numbers $z,w\in \mathbb {C}$ with $w\neq 0$ there is:
${\overline {\left({\frac {z}{w}}\right)}}={\frac {\overline {z}}{\overline {w}}}$
Proof (Conjugation of fractions)
We already know that for the inverse of a complex number $w\in \mathbb {C} \setminus \{0\}$ there is: ${\tfrac {1}{w}}={\tfrac {\overline {w}}{|w|^{2}}}$. We have also seen that real numbers are not changed by the conjugation and that the conjugation is compatible with the multiplication. We first show: ${\overline {\left({\tfrac {1}{w}}\right)}}={\tfrac {1}{\overline {w}}}$. For this we use that ${\tfrac {1}{w|^{2}}}$ is real and that $|w|=|{\overline {w}}|$ holds.
${\begin{aligned}{\overline {\left({\frac {1}{w}}\right)}}&={\overline {\left({\frac {\overline {w}}{|w|^{2}}}\right)}}={\overline {\left({\overline {w}}\cdot {\frac {1}{|w|^{2}}}\right)}}={\overline {\left({\overline {w}}\right)}}\cdot {\overline {\left({\frac {1}{|w|^{2}}}\right)}}\\[0.5em]&={\overline {\left({\overline {w}}\right)}}\cdot {\frac {1}{|w|^{2}}}={\overline {\left({\overline {w}}\right)}}\cdot {\frac {1}{|{\overline {w}}|^{2}}}={\frac {\overline {\left({\overline {w}}\right)}}{|{\overline {w}}|^{2}}}={\frac {1}{\overline {w}}}\end{aligned}}$
This gives us for the conjugation of fractions of complex numbers $z,w\in \mathbb {C}$ with $w\neq 0$:
${\overline {\left({\frac {z}{w}}\right)}}={\overline {\left(z\cdot {\frac {1}{w}}\right)}}={\overline {z}}\cdot {\overline {\left({\frac {1}{w}}\right)}}={\overline {z}}\cdot {\frac {1}{\overline {w}}}={\frac {\overline {z}}{\overline {w}}}$

## Properties of the complex absolute funciton
### Positive definiteness

Theorem (Positive definiteness)
Let $z\in \mathbb {C}$ be a complex number. Then, there is:
$|z|\geq 0\land |z|=0\iff z=0$
Proof (Positive definiteness)
Let $z=a+b\,\mathrm {i}$ be given in Cartesian form. Then, there is $|z|={\sqrt {a^{2}+b^{2}}}\geq 0$. Now we still have to prove the equivalence. For this we show two implications:
Proof step: $z=0\implies |z|=0$
Let $z=0$. Then, there is $a=b=0$. So we have $|z|={\sqrt {a^{2}+b^{2}}}={\sqrt {0+0}}=0$.
Proof step: $|z|=0\implies z=0$
This direction is shown by contradiction. Let $z\neq 0$. This implies $a\neq 0$ or $b\neq 0$. If $a\neq 0$ then, there is $|z|={\sqrt {a^{2}+b^{2}}}\geq {\sqrt {a^{2}}}=|a|>0$. For $b\neq 0$ there is $|z|={\sqrt {a^{2}+b^{2}}}\geq {\sqrt {b^{2}}}=|b|>0$. In every case we have $|z|>0$ and hence $|z|\neq 0$.

### Multiplicativity

Theorem (Multiplicativity)
For $w,z\in \mathbb {C}$ there is $|z\cdot w|=|z|\cdot |w|$.
Proof (Multiplicativity)
Let $w=a+b\,\mathrm {i}$ and $z=c+d\,\mathrm {i} \in \mathbb {C}$. Then, there is
${\begin{aligned}|z\cdot w|^{2}&=|(c+d\,\mathrm {i} )\cdot (a+b\,\mathrm {i} )|^{2}\\&=|(ac-bd)+(cb+da)\,\mathrm {i} |^{2}\\&=(ac-bd)^{2}+(cb+da)^{2}\\&=a^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+d^{2}a^{2}\\&=(c^{2}+d^{2})(a^{2}+b^{2})\\&=|z|^{2}\cdot |w|^{2}\\&=(|z|\cdot |w|)^{2}\end{aligned}}$
Since the bases $|z\cdot w|$ and $|z|\cdot |w|$ of the both squares are non-negative, we are allowed to take the root on both sides. Thus, we get $|z\cdot w|=|z|\cdot |w|$.

### Estimating real and imaginary part

Theorem
For all $z\in \mathbb {C}$ there is $|{\text{Re}}(z)|\leq |z|$ and $|{\text{Im}}(z)|\leq |z|$.
Proof
Let $z=a+b\,\mathrm {i} \in \mathbb {C}$ with $a,b\in \mathbb {R}$. Then with $b^{2}\geq 0$, we have:
$|z|=|a+b\,\mathrm {i} |={\sqrt {a^{2}+b^{2}}}\geq {\sqrt {a^{2}}}=|a|=|\mathrm {Re} (z)|$ Analogously from $a^{2}\geq 0$, we get:
$|z|=|a+b\,\mathrm {i} |={\sqrt {a^{2}+b^{2}}}\geq {\sqrt {b^{2}}}=|b|=|\mathrm {Im} (z)|$

### Triangle inequality

Theorem (Triangle inequality)
For all $w,z\in \mathbb {C}$ there is $|w+z|\leq |w|+|z|$.
Proof (Triangle inequality)
Let $w$ and $z\in \mathbb {C}$. In order to estimate the absolute value, we use the relation $|v|^{2}=v\cdot {\overline {v}}$:
${\begin{aligned}|w+z|^{2}&=(w+z){\overline {(w+z)}}\\&=(w+z)({\overline {w}}+{\overline {z}})\\&=w{\overline {w}}+w{\overline {z}}+{\overline {w}}z+{\overline {z}}z\\&=|w|^{2}+(w{\overline {z}}+{\overline {w{\overline {z}}}})+|z|^{2}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{For }}v\in \mathbb {C} {\text{ there is }}v+{\overline {v}}=2\mathrm {Re} (v)\right.}\\[0.3em]&=|w|^{2}+2\mathrm {Re} (w{\overline {z}})+|z|^{2}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \mathrm {Re} (w{\overline {z}})\leq |\mathrm {Re} (w{\overline {z}})|\leq |w{\overline {z}}|=|w||{\overline {z}}|=|w||z|\right.}\\[0.3em]&\leq |w|^{2}+2|w||z|+|z|^{2}\\&=(|w|+|z|)^{2}\end{aligned}}$

### Estimating the absolute value

Theorem
For all $z\in \mathbb {C}$ there is $|z|\leq |\operatorname {Re} (z)|+|\operatorname {Im} (z)|$.
Proof
First, we show that the square of the inequality holds.
${\begin{aligned}|z|^{2}&=a^{2}+b^{2}=|a|^{2}+|b|^{2}\\&\leq |a|^{2}+2|a||b|+|b|^{2}\\&=(|a|+|b|)^{2}=(|\operatorname {Re} (z)|+|\operatorname {Im} (z)|)^{2}\end{aligned}}$
The two bases $|z|$ and $|\operatorname {Re} (z)|+|\operatorname {Im} (z)|$ are non-negative numbers, so we can take the root on both sides of the inequality. This root preserves inequalities and thus $|z|\leq |\operatorname {Re} (z)|+|\operatorname {Im} (z)|$.

### Inverse triangle inequality

Theorem (Inverse triangle inequality)
For complex numbers $w,z\in \mathbb {C}$ there is ${\big |}|w|-|z|{\big |}\leq |w-z|$.
Proof (Inverse triangle inequality)
In order to prove an inequality $|x|\leq y$, we can also just prove the two inequalities $x\leq y$ and $-x\leq y$. We will apply exactly this technique: in the following, we establish the two inequalities $|w|-|z|\leq |w-z|$ and $|z|-|w|\leq |w-z|$.
Let's start with the first inequality. We use the triangle inequality of the complex amount and the trick of "inserting a fruitful $0$" (which often appears together with triangle inequalities):
$|w|=|(w-z)+z|\leq |w-z|+|z|$
By transformation we obtain $|w|-|z|\leq |w-z|$. We show the second inequality analogously, where the roles of $z$ and $w$ are reversed. In addition, we gradually transform $|z-w|$ into $|w-z|$:
${\begin{aligned}|z|&=|(z-w)+w|\leq |z-w|+|w|\\&=|(-1)(w-z)|+|w|\\&=|(-1)||w-z|+|w|\\&=|w-z|+|w|\end{aligned}}$
This implies $|z|-|w|\leq |w-z|$. So in all, we have proven the two inequalities $|w|-|z|\leq |w-z|$ and $-(|w|-|z|)\leq |w-z|$. This establishes $||w|-|z||\leq |w-z|$.
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