The properties of complex numbers

Moduli of the opposite and the conjugate

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

And finally,

Modulus of a product

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

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:

Modulus of a complex number raised to an integer power

Let \(z \in \mathbb{C}\).

Let us prove by induction that for any natural integer \(n \in \mathbb{N}\), the following property holds:

1. Initialization (for \(n = 0\))

   By convention, for any complex number \(z \neq 0\), we have \(z^0 = 1\). Let us calculate both sides:

   $$ \begin{cases} |z^0| = |1| = 1 \\ |z|^0 = 1 \end{cases} $$

   The equality holds, so the statement \((P_0)\) is true.

2. Inductive Step

   Let \(k \in \mathbb{N}\) be a fixed natural integer. Suppose that the statement \((P_k)\) is true, i.e., that \(|z^k| = |z|^k\). Let us check if it remains true at rank \(k+1\), that is:

   $$ \left|z^{k+1}\right| = |z|^{k+1} \qquad (P_{k+1}) $$

   By the laws of exponents, we know that \(z^{k+1} = z^k \times z\). Applying the modulus, we obtain:

   $$ |z^{k+1}| = |z^k \times z| $$

   Now, we have previously proved that the modulus of a product is equal to the product of the moduli:

   $$ \forall (z, z') \in \mathbb{C}^2, \enspace |z z'| = |z| \cdot |z'| $$

   This allows us to separate the product:

   $$ |z^{k+1}| = |z^k| \cdot |z| $$

   Using our inductive hypothesis (\(|z^k| = |z|^k\)), we can substitute this term:

   $$ |z^{k+1}| = |z|^k \cdot |z| = |z|^{k+1} $$

   The property is therefore inductive.

3. Conclusion

   The statement \((P_n)\) is true at rank \(0\) and is inductive for any natural integer \(n\). By the principle of mathematical induction, we conclude that:

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

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.

   $$\overline{z} = |z| \cdot \left( \cos(\theta) - i \hspace{0.2em} \sin(\theta) \right) $$

   But,

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

   So,

   $$\overline{z} = |z| \cdot \left( \cos(-\theta) + i \sin(-\theta) \right) $$

   Hence,

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

2. For the opposite \( -z \)

   $$-z = -|z| \cdot \left( \cos(\theta) + i \hspace{0.2em} \sin(\theta) \right) $$
   $$-z = |z| \cdot \left( -\cos(\theta) - i \hspace{0.2em} \sin(\theta) \right) $$

   But,

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

   So,

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

   Hence,

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

And as a result,

Argument of a product

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

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

So,

So that,

And as a result,

Argument of an inverse

Let \(z \in \hspace{0.04em}\mathbb{C}^*\) 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 \(z' \in \hspace{0.04em}\mathbb{C}^*\) a non-zero complex number.

Let us write the quotient \(\frac{z}{z'}\) in 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\(: z^2 \)

   $$ z^2 = \Bigl[ |z| \cdot \left(\cos(\theta) + i \hspace{0.2em} \sin(\theta) \right) \Bigr]^2 $$

   Thanks to the properties of the modulus 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:

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

   1. Calculating the first term

      $$ z = |z| \cdot \left(\cos(\theta) + i \hspace{0.2em} \sin(\theta) \right) $$
      $$ z^0 = |z|^0.\left(\cos(0 \times \theta) + isin( 0 \times \theta) \right) $$
      $$ 1 = 1 \times ( 1 + 0) $$

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

   2. Inductive step

      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 \(z^{k+1}\):

         $$ z^{k+1} = |z|^{k+1}.\left(\cos(\theta) + i \hspace{0.2em} \sin(\theta) \right)^{k+1} $$

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

         $$ \forall z, z' \in \hspace{0.04em} \mathbb{C}^2, $$

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

         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,

Conjugate of a sum

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

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.

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.

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.

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,