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Complex numbers

The reason for the introduction of complex numbers is so that every quadratic equation will have a solution. For instance, an equation x^2-1 = 0 contains solutions in a set of real numbers, however x^2+1=0 does not contains solutions in a set of real numbers. Because of these and similar equations, we expand the set of real numbers (\mathbb{R}) to the set in which they will have the solution.

Let i be the intended solution to the equation x^2 + 1 =0; therefore i^2 = -1.  The number i is called the unit imaginary number. The unit imaginary number has the main role in describing a set of complex numbers \mathbb{C} which will be the extension of a set of real numbers \mathbb{R}.

The product of any real number y and imaginary unit i is a complex number. Numbers such as these are called imaginary numbers.

A complex number is the addition of a real and an imaginary number, that is, a complex number z is the number of the shape z= x + yi, where x and y are real numbers. The number x is called a real part, and y is called an imaginary part of the complex number z. We write:

    \[x = Re z, \quad \quad  y= Im z.\]

A set of complex numbers is denoted as:

    \[\mathbb{C} = \{x + yi : x, y \in \mathbb{R} \}.\]

Two complex numbers z and w are equal if

    \[z=w \Leftrightarrow Re z = Re w, Im z = Im w.\]

Example 1. Determine x and y such that the following is valid:

    \[(2 +3i) + (x+yi) = -7 +3i.\]

Solution:

    \[(2 +3i) + (x+yi) = -7 +3i\]

    \[(2+x) + (3+y)i  = -7 +3i\]

Two complex numbers are equal iff their real and imaginary parts are equal. Therefore, we have:

    \[2+x = -7 \Rightarrow x = -7 -2 = -9\]

and

    \[3+y = 3 \Rightarrow y= 3-3 =0.\]

Powers of the imaginary unit 

    \[i^0 = 1\]

    \[i^1 = i\]

    \[i^2 = -1\]

    \[i^3 = i^2 \cdot i = -1 \cdot i = -i\]

    \[i^4 = i^3 \cdot i = -i \cdot i = -i^2 = 1.\]

 

If we count further

    \[i^5 = i^4 \cdot i = 1 \cdot i = i\]

    \[i^6 = i^5 \cdot i = i\cdot i = i^2 = -1\]

    \[i^7 = i^6 \cdot i = -1 \cdot i = -i\]

    \[i^8 = i^7 \cdot i = -i \cdot i = -i^2 = 1\]

we can observe that values of powers are repeated. Therefore,

    \[i^n = i ^{4a + b} = i^{4a} \cdot i ^b = 1 \cdot i^b = i^b \quad b\in\{0,1,2,3\}, a \in \mathbb{Z}\]

is valid.

Example 2.

Calculate the following:

    \[(-2i^{1023} - 3i^{343}) ( -7i^{234} + i^{456}).\]

Solution:

1023 by dividing with 4 gives the rest 3, that is

    \[i^{1023} = i^{4 \cdot 255} \cdot i ^3 = 1 \cdot (-i) = -i.\]

Similarly, we obtain:

    \[i^{343} = i^{4 \cdot 85 } \cdot i^3 = 1 \cdot (-i) = -i,\]

    \[i^{234} = i^{4 \cdot 58 } \cdot i^2 = 1 \cdot (-1) =-1,\]

    \[i^{456} = i^{4 \cdot 114} = 1.\]

Finally, we have:

    \[\begin{aligned} (-2i^{1023} - 3i^{343}) ( -7i^{234} + i^{456}) &=( -2 \cdot (-i) - 3 \cdot (-i)) (-7 \cdot(-1) + 1) \\ & = (2i + 3i)(7+1)\\ &= 5i \cdot 8 \\ &= 40i. \\ \end{aligned}\]

Example 3. Calculate:

    \[i + i ^2 + i ^3 + \ldots + i^{10}.\]

Solution:

Since

    \[i + i^2 + i^3 + i^4 = i + (-1) - i + 1 = 0\]

then we have

    \[\begin{aligned} \underbrace{i + i^2 + i^3 + i^4 }_{=0} + \underbrace{i^5 + i^6 + i^7 + i^8 }_{=0} + i^9 +i^{10} &= 0 + 0 + i^9 +i^{10} \\ &= i^{4 \cdot 2} \cdot i + i^{4 \cdot 2} \cdot i^{2} \\ &= 1 \cdot i + 1 \cdot (-1) \\ & = 1 -1 \\ & =0. \\ \end{aligned}\]

 

 

 

 

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