The properties of complex numbers

Modules\(:|z|\)

We write \( |z| \) the module of a complex number \( z \).

Let be \( (x, y) \in \hspace{0.05em} \mathbb{R}^2, \enspace z \in \mathbb{C}, \enspace \Biggl \{ \begin{gather*} z = x + iy \\ |z| = \sqrt{x^2 + y^2 } \end{gather*} \)

Modules of the opposite and the conjugate

Module of a product

In the same way, we will have:

Module of a complex number raised to an integer power

Arguments\(: arg(z) \)

Let be \( (x, y) \in \hspace{0.05em} \mathbb{R}^2, \enspace z \in \mathbb{C}, \)

$$ \Biggl \{ \begin{gather*} z = x + iy \\ z = |z|.\left(cos(\theta) + isin(\theta) \right) \end{gather*} $$

We write \( arg(z) \) the argument of a cmplex number \( z \).

Arguments of conjugate and opposite

$$ \Biggl \{ \begin{gather*} arg(\overline{z}) = -arg(z) \\ arg( -z) = \pi + arg(z) \end{gather*} $$

Argument of a product

Argument of an inverse

Argument of a quotient

Argument of a complex number raised to an integer power

Conjugates\(: \overline{z}\)

We write \( \overline{z} \) the conjugate of a complex number \( z \).

Let be \( (x, y) \in \hspace{0.05em} \mathbb{R}^2, \enspace z \in \mathbb{C}, \enspace \Biggl \{ \begin{gather*} z = x + iy \\ \overline{z} = x -iy \end{gather*} \)

Conjugate of a sum

In the same way,

Conjugate of a product

Conjugate of a quotient

Complex number multiplied by its conjugate

Conjugate of a complex number raised to an integer power

Recap table of the properties of the complex numbers formulas

Click on the title to access to the recap table.

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Demonstrations

Modules\(: |z|\)

Modules of the opposite and the conjugate

Let us write the complex numbers \( |-z|\) et \( | \overline{z} | \) under their algebraic form.

$$ \Biggl \{ \begin{gather*} |-z| = \sqrt{(-x)^2 + (-y)^2} = \sqrt{x^2 + y^2 } = |z| \\ | \overline{z} | = \sqrt{x' + (-y)^2 } = \sqrt{x^2 + y^2 }= |z| \end{gather*} $$

And finally,

Module of a product

Let us write the complex numbers \( z, z' \) under their algebraic form.

$$ \Biggl \{ \begin{gather*} z = x + iy \\ z' = x' + iy' \end{gather*} $$

Calculating \( z z' \), we do have:

Then, calculating \( | z z' | \):

In the end, calculating \( | z| \hspace{0.2em}. |z' | \) we do have:

We now notice that both expressions \( (1) \) and \( (2) \) are equals, so:

In the same way, we will have:

Module of a complex number raised to an integer power

Let us write the complex number \(z\) under its algebraic form.

Let us no calculate \(z^n\):

Then,

But,

We definitely have,

Arguments\(: arg(z) \)

Arguments of conjugate and opposite

Let us write the complex number \(z\) under its trigonometric form.

1. For the conjugate \(\overline{z}\)

.

Let us write the complex number \( \overline{z} \) under its trigonometric form.

$$\overline{z} = |z|.\left( cos(\theta) - isin(\theta) \right) $$

But,

$$ \Biggl \{ \begin{gather*} cos(\theta) = cos(-\theta) \\ -sin(\theta) = sin(-\theta) \end{gather*} $$

So,

$$\overline{z} = |z|.\left( cos(-\theta) + isin(-\theta) \right) $$

Hence,

$$ arg(\overline{z}) = -arg(z) $$

2. For the opposite \( -z \)

$$-z = -|z|.\left( cos(\theta) + isin(\theta) \right) $$
$$-z = |z|.\left( -cos(\theta) - isin(\theta) \right) $$

But,

$$ \Biggl \{ \begin{gather*} -cos(\theta) = cos(\pi + \theta) \\ -sin(\theta) = sin(\pi + \theta) \end{gather*} $$

So,

$$ -z = |z|.\left( cos(\pi + \theta) + isin(\pi + \theta) \right) $$

Hence,

$$ arg(-z) = \pi +arg(z) $$

And as a result,

$$ \Biggl \{ \begin{gather*} arg(\overline{z}) = -arg(z) \\ arg( -z) = \pi + arg(z) \end{gather*} $$

Argument of a product

Let us write the complex numbers \( z, z' \) under their trigonometric form.

$$ \Biggl \{ \begin{gather*} z = |z|.\left(cos(\theta) + isin(\theta) \right) \\ z' = |z'|.\left(cos(\theta') + isin(\theta') \right) \end{gather*} $$

Performing the product \( z z' \), we do obtain:

So,

So that,

And as a result,

Argument of an inverse

Let \(z \in \hspace{0.05em}\mathbb{C}^*\) be a non-zero complex number.

Let us start fro mthe following equation:

Then,

So that,

And finally,

Argument of a quotient

Let \(z \in \mathbb{C}\) be a complex number and \(z' \in \hspace{0.05em}\mathbb{C}^*\) a non-zero complex number.

Let us write the quotient \(\frac{z}{z'}\) under the form of a product:

So that,

So,

And as a result,

Argument of a complex number raised to an integer power

Let us write the complex number \( z \) under its trigonometric form.

1. Calculating the square\(: z^2 \)

$$ z^2 = \Bigl[ |z|.\left(cos(\theta) + isin(\theta) \right) \Bigr]^2 $$

Thanks to the properties of the module raised to an integer power \( | z^n | \), we know that:

$$ \forall z \in \mathbb{C}, \enspace \forall n \in \mathbb{N}, $$

$$ |z^n| = |z|^n $$

So,

$$ z^2 = |z|^2.\left(cos(\theta) + i.sin(\theta) \right)^2$$
$$ z^2 = |z|^2.\left(cos^2(\theta) + 2i.sin(\theta)cos(\theta) - sin^2(\theta) \right)$$
$$ z^2 = |z|^2.\left(cos^2(\theta) - sin^2(\theta) + 2i.sin(\theta)cos(\theta) \right)$$

Now, thanks to trigonometric duplicaiton formulas, we know that:

$$ \forall \alpha \in \mathbb{R}, \enspace \Biggl \{ \begin{gather*} sin(2\alpha) = 2 sin(\alpha) cos(\alpha) \\ cos(2\alpha) = cos^2(\alpha) - sin^2(\alpha) \end{gather*} $$

Thus, we identify that:

$$ z^2 = |z|^2.\left(cos(2\theta) + i .sin(2\theta) \right)$$

And,

$$ arg(z^2) = 2 . arg(z) $$



2. Proof by a recurrence

Let us show by a recurrence that:

$$ \forall z \in \mathbb{C}, \enspace \forall n \in \mathbb{Z},$$
$$ arg(z^n) = n . arg(z) \qquad (S_n) $$

1. Calculating the first term

$$ z = |z|.\left(cos(\theta) + isin(\theta) \right) $$
$$ z^0 = |z|^0.\left(cos(0 \times \theta) + isin( 0 \times \theta) \right) $$
$$ 1 = 1 \times ( 1 + 0) $$

Thus, \((S_0)\) is true.




2. Heredity

1. With an exponent \( n \) in the natural numbers set \( (n \in \mathbb{N}) \)

Let \( k \in \mathbb{N} \) be a nutral number.

Let us assume that the statement \((S_k)\) is true for all \( k \), and let us verify if it is also the case for \((S_{k + 1})\).

If it is so, we should obtain as a result that:

$$ arg(z^{k+1}) = (k+1) . arg(z) \qquad (S_{k + 1}) $$



Consequently, let us calculate \(z^{k+1}\):

$$ z^{k+1} = |z|^{k+1}.\left(cos(\theta) + isin(\theta) \right)^{n+1} $$

But, we know that the argument of a product is the sum of the product's factors of this product:

$$ \forall z, z' \in \hspace{0.05em} \mathbb{C}^2, $$

$$ arg( z z') = arg(z) + arg(z') $$

So in our case that,

$$ arg(z^{k+1}) = arg( z . z^k) = arg(z) + arg(z^k) $$
$$ arg(z^{k+1}) = arg(z) + arg(z) + arg(z^{k-1}) $$

And so on until:

$$ arg\left(z^{k+1}\right) = (k+1).arg(z) $$

Thus, \((S_{k + 1})\) is true in the natural numbers set \( \mathbb{N}\).

Let us now proove it backwards.



2. With an exponent \( n \) in the natural integers set \( (n \in \mathbb{Z}) \)

Let \( k \in \mathbb{Z} \) be an integer.

Let us assume that the statement \((S_k)\) is true for all \( k \), and let us verify if it is also the case for \((P_{k - 1})\).

If it is so, we should obtain as a result that:

$$ arg(z^{k-1}) = (k-1) . arg(z) \qquad (P_{k - 1}) $$



Calculating now \(z^{k-1}\), we do have:

$$ z^{k-1} = \frac{z^k}{z} $$

But, we know that the argument of a quotient of two complex numbers is the difference of arguments of these numbers:

$$ \forall z \in \mathbb{C}, \enspace \forall z' \in \mathbb{C^*},$$

$$ arg\left(\frac{z}{z'}\right) = arg(z) -arg(z') $$

So that,

$$ arg(z^{k-1}) = arg\left( \frac{z^k}{z} \right) = arg(z^k) - arg(z) $$
$$ arg(z^{k-1}) = k.arg(z) - arg(z) $$
$$ arg(z^{k-1}) = (k-1).arg(z) $$

Thus, \((P_{k - 1})\) is true for the integers set \( \mathbb{Z}\).




3. Conclusion

The statement \((S_n)\) is true for its first term \(n_0 = 0\) and it is hereditary from terms to terms for all \(n \in \mathbb{Z}\), increasingly and decreasingly.

Thus, by the recurrence principle, this is true for all \(n \in \mathbb{Z}\).

And finally, as a result we do have,

Conjugates\(: \overline{z}\)

Conjugate of a sum

Let us write the complex numbers \( z_1, z_2 \) under their algebraic form.

$$ \Biggl \{ \begin{gather*} z_1 = x_1 + iy_1 \\ z_2 = x_2 + iy_2 \end{gather*} $$

Performing their sum, we do have:

Now, applying the conjugate of it:

And finally,

In the same way,

Conjugate of a product

Let us write the complex numbers \( z_1, z_2 \) under their algebraic form.

$$ \Biggl \{ \begin{gather*} z_1 = x_1 + iy_1 \\ z_2 = x_2 + iy_2 \end{gather*} $$

Performing their product, we do have:

Now, applying the conjugate of it:

Let us now calculate the product \( \overline{z_1}. \overline{z_2} \) separately:

After having calculated both expressions \( (3) \) and \( (4) \), we notice that they are equals:

As a result we do have,

Conjugate of a quotient

Let us write the complex numbers \( z \) and \( \overset{-}{z} \) under their algebraic form.

$$ \Biggl \{ \begin{gather*} z= x + iy \\ \overset{-}{z} = x - iy \end{gather*} $$

Performing their quotient, we do have:

Multiplying both numerator and denominator by the denominator's conjugate,

So in our case,

Then, developping the numerator,

Now, applying the conjugate of it:

Let us now calculate the quotient \( \frac{\overline{z_1}}{ \overline{z_2}} \) separately:

In the same way as as above,

After having calculated both expressions \( (5) \) and \( (6) \), we notice that they are equals:

And finally,

Complex number multiplied by its conjugate

Let us write the complex numbers \( z \) and \( \overset{-}{z} \) under their algebraic form.

$$ \Biggl \{ \begin{gather*} z= x + iy \\ \overset{-}{z} = x - iy \end{gather*} $$

Let us calculate their product:

So in our case,

And finally,

Conjugate of a complex number raised to an integer power

Let us write the complex numbers \( z \) under its trigonometric form.

Calculating \( z^n \), we do have:

Then, we can now write that:

Let us now apply the conjugate of it:

Let us now calculate \( (\overline{z})^n \) seperately, starting from \( \overline{z} \).

With the Moivre's formula again, we do have:

After having calculated both expressions \( (7) \) et \( (8) \), we notice that they are equals:

And as a result,

Recap table of the properties of the complex numbers formulas