# How are imaginary numbers useful?

## Complex numbers

Complex numbers (symbol: \$ z \$) represent an extension of the number range. This extension is necessary to be able to solve equations such as \$ x ^ 2 = -1 \$. For this equation we cannot find a real number from \$ \ mathbb {R} \$ that would solve this equation.

Complex numbers can be represented in the form \$ z = a + b \ cdot i \$. The part \$ a + b \$ of the function is called the real part and the \$ i \$ imaginary part.

### Basics of complex numbers

To get started, watch the tutorial on complex numbers

Complex numbers, overview, imaginary unit, real part, imaginary part | Math by Daniel Jung Complex numbers are due to their construction on the complex number plane arranged. A complex number is made up of the following parts:
\$ \ quad z = a + bi \$

Real part Re (z) and imaginary part Im (z):
\$ \ quad \$ Re (z) = \$ a \ in \ mathbb {R} \$
\$ \ quad \$ Im (z) = \$ b \ in \ mathbb {R} \ quad \$ Note: Im (z) \$ \ neq bi \$!

Amount | z | and argument \$ \ varphi \$:
\$ \ quad | z | = r = \ sqrt {(\ textrm {Re {z}}) ^ 2 + (\ textrm {Im {z}}) ^ 2} = \ sqrt {a ^ 2 + b ^ 2} \
\ quad \ varphi = \ arg (z) \ quad \ textrm {angle between pos. Re-axis and line} \ \ \ overline {0z} \$

For any complex number \$ z \$ there are complex conjugates \$ \ overline {z} \$:
\$ \ quad \ overline {z} = a - bi \$

Imaginary unit \$ i \$:
\$ \ quad i ^ 2 = -1 \$

Contrary to many spellings and internet sources, \$ i \$ is purely mathematically \$ \ textit {not} \$ to be equated with \$ \ sqrt {-1} \$.

The following calculation shows the conflict in the assumption that there would be \$ \ sqrt {-1} \$ and that could be expected. cf. \$ \ ast ^ 2 \$: The "\$ = \$" would also not be correct without the assumption of the existence of \$ \ sqrt {-1} \$! From \$ \ ast ^ 1 \$ the calculation goes wrong and a contradiction arises:

\ begin {align *}
\ quad -1 = i ^ 2 = i \ cdot i = ^ {\ ast ^ 1} \ \ sqrt {-1} \ cdot \ sqrt {-1} = ^ {\ ast ^ 2} \ \ sqrt {(- 1) (- 1)} = \ sqrt {1} = 1 \ quad \ Longrightarrow -1 \ neq 1
\ end {align *}

Useful to know
\ begin {align *}
& x ^ 2 = -4 \ \ Rightarrow \ x = \ pm \ sqrt {-4} = \ pm \ sqrt {4} \ cdot \ sqrt {-1} = \ pm2i && \ quad \ text {unclean path!} \ [2mm]
& x ^ 2 = -4 \ \ Rightarrow \ x = \ pm \ sqrt {-4} = \ pm \ sqrt {4i ^ 2} = \ pm2i && \ quad \ text {slightly better way!} \ [2mm]
& x ^ 2 = -4 \ \ Rightarrow \ x ^ 2 = 4i ^ 2 \ Rightarrow \ x = \ pm2i && \ quad \ text {correct way!}
\ end {align *}

### Forms of representation of complex numbers

Complex numbers, Euler's identity, polar form, math help online | Math by Daniel Jung Since the complex numbers are on one level, we use polar coordinates for a clear assignment of the numbers. This allows the numbers to be converted into the \$ \ textit {polar form} \$. In many calculations this representation has advantages over the classic \$ \ textit {Cartesian representation} \$ of numbers. As we can see from the graphic, a complex number is defined either by its real and imaginary part or by "amount and argument":

\$ \ quad \ text {Cartesian} z = a + bi \$
\$ \ quad \ text {Polar} z = r \ exp ^ {\ varphi i} \$

It makes sense that the polar form is described by the e-function, since the e-function in the complex depends on the magnitude and the argument. If we reduce these to real numbers, the usual exponential function results again, as already known from school.

Useful to know

There is also the so-called trigonometric representation. However, this is immediately apparent from the polar form and formula. Therefore we do not list them separately. ### Convert representations

Cartesian to polar form

\$ r = \ sqrt {\ textrm {Re {z}} ^ 2+ \ textrm {Im {z}} ^ 2} \$
\$ \ varphi = \ begin {cases} \ arccos \ left (\ frac {\ textrm {Re {z}}} {r} \ right) & \ text {for} \ textrm {Im {z}} \ geq 0 \ \ 2 \ pi- \ arccos \ left (\ frac {\ textrm {Re {z}}} {r} \ right) & \ text {for} \ textrm {Im {z}} <0 \ end {cases} \$

Useful to know

Phi can also be calculated using \$ \ arcsin (\ ldots) \$ or \$ \ arctan \$, but with \$ \ arccos \$ we don't have to 1. look at the quadrant in which the number lies in the complex number plane and 2. are normally also the polar coordinates are defined in \$ \ mathbb {R} ^ 2 \$.

Polar form to Cartesian

\$ \ quad \ textrm {Re {z}} = r \ cdot \ cos (\ varphi) \$
\$ \ quad \ textrm {Im {z}} = r \ cdot \ sin (\ varphi) \$

### Examples of converting complex numbers

From Cartesian to Polar:
\ begin {align *}
z & = 3 + \ sqrt {3} \ cdot i \ qquad \ text {with} \
& \ qquad r = \ sqrt {9 + 3} = \ sqrt {12} \
& \ qquad \ varphi \ stackrel {\ sqrt {3}> 0} {=} \ arccos \ left (\ frac {3} {\ sqrt {12}} \ right) = \ arccos \ left (\ frac {\ sqrt {3 ^ 2}} {\ sqrt {3 \ cdot 4}} \ right) = \ arccos \ left (\ frac {\ sqrt {3}} {2} \ right) = \ frac {\ pi} {6} \
z & = 2 \ sqrt {3} \ cdot \ exp ^ {\ frac {\ pi} {6} \ cdot i}
\ end {align *}

From Polar to Cartesian:
\ begin {align *}
z & = 2 \ cdot \ exp ^ {\ frac {5} {4} \ pi i} \ qquad \ text {with} \
& \ qquad \ textrm {Re {z}} = 2 \ cos \ left (\ frac {5} {4} \ pi \ right) = 2 \ cos \ left (\ pi + \ frac {1} {4} \ pi \ right) = 2 \ Bigl (- \ cos \ left (\ frac {1} {4} \ pi \ right) \ Bigr) = - 2 \ frac {1} {\ sqrt {2}} \
& \ qquad \ textrm {Im {z}} = 2 \ sin \ left (\ frac {5} {4} \ pi \ right) = 2 \ sin \ left (\ pi + \ frac {1} {4} \ pi \ right) = 2 \ Bigl (- \ sin \ left (\ frac {1} {4} \ pi \ right) \ Bigr) = - 2 \ frac {1} {\ sqrt {2}} \
z & = \ sqrt {2} + \ sqrt {2} \ cdot i
\ end {align *} ### Overview of calculation rules

All the arithmetic laws for numbers from \$ \ mathbb {C} \$ apply, as we know them for the numbers from \$ \ mathbb {R} \$. You just have to be careful not to mix real and imaginary parts in calculations. You can therefore view the \$ i \$ like a variable.

Calculating with complex numbers, sum, difference, product | Math by Daniel Jung The following applies to Cartesian representation ...
\$ \ quad z_1 = a + bi \ textrm {and} z_2 = c + di \ textrm {and} n \ in \ mathbb {N} \$

\$ \ quad z_1 + z_2 = (a + c) + (b + d) i \$

Multiplication of complex numbers
\$ \ quad z_1 \ cdot z_2 = (ac-bd) + (ad + bc) i \$

Division of complex numbers
\$ \ quad \ frac {z_1} {z_2} = \ frac {z_1 \ overline {z_2}} {z_2 \ overline {z_2}} = \ frac {(a + bi) (c-di)} {c ^ 2 + d ^ 2} = \ frac {ac + bd} {c ^ 2 + d ^ 2} + \ frac {bc-ad} {c ^ 2 + d ^ 2} i \$

Powering complex numbers to the power
\$ \ quad z_1 ^ n = \ ldots \ text {not recommended from} n \ geq 5 \$
\$ \ quad z_1 ^ {z_2} = \ ldots \ text {not recommended} \$

Extract the square root of a complex number
\$ \ quad \ sqrt [n] {z_1} = \ ldots \ text {not recommended} \$

### The following applies to polar form ...

\$ \ quad z_1 = r_1 \ cdot \ exp ^ {\ varphi_1i} \ textrm {and} z_2 = r_2 \ cdot \ exp ^ {\ varphi_2i} \ textrm {and} n \ in \ mathbb {N} \$
\$ \ quad z_1 + z_2 = \ ldots \ text {not recommended} \$

product
\$ \ quad z_1 \ cdot z_2 = (r_1 \ cdot r_2) \ cdot \ exp ^ {(\ varphi_1 + \ varphi_2) i} \$

division
\$ \ quad \ frac {z_1} {z_2} = \ frac {r_1 \ cdot \ exp ^ {\ varphi_1i}} {r_2 \ cdot \ exp ^ {\ varphi_2i}} = \ frac {r_1} {r_2} \ cdot \ exp ^ {(\ varphi_1- \ varphi_2) i} \$

Potentiate
\$ \ quad z_1 ^ n = \ left (r_1 \ cdot \ exp ^ {\ varphi_1i} \ right) ^ n = r_1 ^ n \ cdot \ exp ^ {n \ varphi_1i} \$
\$ \ quad z_1 ^ {z_2} = \ exp ^ {z_2 \ cdot \ ln (z_1)} \ quad \ text {with} \ quad z_2 \ cdot \ ln (z_1) = r_2 \ exp ^ {\ varphi_2i} \ cdot \ bigl (\ ln (r_1) + \ varphi_1i \ bigr) \$
\$ \ quad \ quad \ text {then} z_2 \ textrm {convert into Cartesian representation} \$

Take root
\$ \ quad \ sqrt [n] {z_1} = \ ldots \ text {use product equation} \$

In general, a parallel to the vector calculation in \$ \ mathbb {R} ^ 2 \$ can be seen in certain arithmetic operations: When complex numbers are multiplied (divided), the amounts are multiplied (divided) and the arguments added (subtracted).

### Selection of complex functions and complex roots

Selected functions:

Complex exponential function
\$ \ quad \ exp ^ {z} = \ exp ^ {\ Re {z}} \ Bigl (\ cos \ bigl (\ textrm {Im {z}} \ bigr) + i \ sin \ bigl (\ textrm {Im {z}} \ bigr) \ Bigr) \ qquad \ text {so} \$ also applies
\$ \ quad r \ cdot \ exp ^ {\ varphi i} = r \ cdot \ Bigl (\ cos \ bigl (\ varphi \ bigr) + i \ sin \ bigl (\ varphi \ bigr) \ Bigr) \$

Complex logarithm (for z in polar form applies)
\$ \ quad \ ln {(z)} = \ ln {(r \ exp ^ {\ varphi i})} = \ ln (r) + \ varphi i \$

Complex numbers, determining the ln of z, online math help, explanatory video | Math by Daniel Jung Complex trigonometric functions
\$ \ quad \ sin (z) = \ frac {1} {2i} \ left (\ exp ^ {iz} - \ exp ^ {- iz} \ right) \$
\$ \ quad \ cos (z) = \ frac {1} {2} \ left (\ exp ^ {iz} + \ exp ^ {- iz} \ right) \$

Complex root extraction (for z in polar form applies)
\$ \ quad \ sqrt [n] {z} = \ sqrt [n] {r} \ cdot \ text {exp} \ left \ {\ frac {\ varphi + 2k \ pi} {n} \ cdot i \ right \ }, \ quad \ forall \, k \ in \ {0,1, \ ldots, n-1 \} \$

Danger!
Will the n-th root taken from a complex number, we get alwaysn Solutions (knowledge from the fundamental theorem of algebra). This is fundamentally different from pulling roots in the real.

### Example of “complex root extraction” for z in polar form

The formula for complex root extraction looks complicated. That is why we provide you with an example calculation including a sketch:

Find all the solutions to the equation \$ z ^ 5 = 3-3 \ sqrt {3} \$.

First we convert again from Cartesian to polar:

\ begin {alignat *} {2}
& 3-3 \ sqrt {3} \ cdot i \ qquad \ text {with} \
& \ qquad r = \ sqrt {9 + 9 \ cdot 3} = \ sqrt {36} = 6 \
& \ qquad \ varphi \ stackrel {-3 \ sqrt {3} <0} {=} 2 \ pi- \ arccos \ left (\ frac {3} {6} \ right) = 2 \ pi- \ frac {\ pi} {3} = \ frac {5} {3} \ pi \
\ rightarrow \ & 3-3 \ sqrt {3} \ cdot i \ = 6 \ cdot \ exp ^ {\ frac {5} {3} \ pi \ cdot i}
\ end {alignat *}
Hence for z:
\ begin {align *}
\ Rightarrow \ z & = \ sqrt  {z_1} = \ sqrt  {6} \ cdot \ text {exp} \ left \ {\ frac {\ frac {5} {3} \ pi + 2k \ pi } {5} \ cdot i \ right \} = \ sqrt  {6} \ cdot \ text {exp} \ left \ {\ biggl (\ frac {1} {3} \ pi + \ frac {2} { 5} k \ pi \ biggr) \ cdot i \ right \} \
& = \ sqrt  {6} \ cdot \ text {exp} \ left \ {\ biggl (\ frac {5} {15} \ pi + \ frac {6} {15} k \ pi \ biggr) \ cdot i \ right \}, \ quad \ forall \, k \ in \ {0,1,2,3,4 \}
\ end {align *}
So the solutions are:

\ begin {alignat *} {3}
& z_1 && = \ sqrt  {6} \ cdot \ exp ^ {\ frac {1} {3} \ cdot \ pi \ cdot i} && \ qquad (k = 0) \
& z_2 && = \ sqrt  {6} \ cdot \ exp ^ {\ frac {11} {15} \ cdot \ pi \ cdot i} && \ qquad (k = 1) \
& z_3 && = \ sqrt  {6} \ cdot \ exp ^ {\ frac {17} {15} \ cdot \ pi \ cdot i} && \ qquad (k = 2) \
& z_4 && = \ sqrt  {6} \ cdot \ exp ^ {\ frac {23} {15} \ cdot \ pi \ cdot i} && \ qquad (k = 3) \
& z_5 && = \ sqrt  {6} \ cdot \ exp ^ {\ frac {29} {15} \ cdot \ pi \ cdot i} && \ qquad (k = 4)
\ end {alignat *}

All solutions of \$ z ^ 5 = 6-3 \ sqrt {3} \ cdot i \$ in the complex number plane: In theory, it is sufficient to calculate \$ z_1 \$ and construct \$ z_ {2, \ ldots, 5} \$ geometrically. All solutions \$ z_1 \ textrm {to} z_5 \$ are the corner points of a regular polygon on the circle with radius \$ \ sqrt  {6} \$. This arrangement allows us to determine the remaining four solutions besides \$ z_1 \$. The angle (\$ \ frac {2 \ pi} {5} \$) between all solutions is added up - with \$ z_1 \$ as the "initial solution".

So:

\ begin {alignat *} {3}
& z ^ 5 && = 3-3 \ sqrt {3} \
\ Leftrightarrow \ & z ^ 5 && = 6 \ cdot \ exp ^ {\ frac {5} {3} \ pi \ cdot i} \
& z_1 && = \ sqrt  {6} \ cdot \ exp ^ {\ frac {1} {5} \ cdot \ frac {5} {3} \ pi \ cdot i} = \ sqrt  {6} \ cdot \ exp ^ {\ frac {1} {3} \ pi \ cdot i} \ [4mm]
& && \ text {da 5th root:} \ frac {2 \ pi} {5} \ text {from one to the next solution to the argument add.:}\[4mm]
\ Rightarrow \ & z_2 && = \ sqrt  {6} \ cdot \ exp ^ {\ left (\ frac {1} {3} \ pi + \ frac {2 \ pi} {5} \ right) \ cdot i} = \ sqrt  {6} \ cdot \ exp ^ {\ frac {11} {15} \ pi \ cdot i} \
\ Rightarrow \ & z_3 && = \ sqrt  {6} \ cdot \ exp ^ {\ left (\ frac {11} {15} \ pi + \ frac {2 \ pi} {5} \ right) \ cdot i} = \ sqrt  {6} \ cdot \ exp ^ {\ frac {17} {15} \ pi \ cdot i} \
\ Rightarrow \ & z_4 && = \ sqrt  {6} \ cdot \ exp ^ {\ left (\ frac {17} {15} \ pi + \ frac {2 \ pi} {5} \ right) \ cdot i} = \ sqrt  {6} \ cdot \ exp ^ {\ frac {23} {15} \ pi \ cdot i} \
\ Rightarrow \ & z_5 && = \ sqrt  {6} \ cdot \ exp ^ {\ left (\ frac {23} {15} \ pi + \ frac {2 \ pi} {5} \ right) \ cdot i} = \ sqrt  {6} \ cdot \ exp ^ {\ frac {29} {15} \ pi \ cdot i}
\ end {alignat *}

Whether you memorize the formula or calculate according to this scheme is of course up to you.

### Powers of i and combined example \ begin {align *}
& \ begin {array} {* 4 {> {\ displaystyle} l}}
i ^ 2 = -1 & \ quad, \, i ^ 3 = -i & \ quad, \, i ^ 4 = (- 1) ^ 2 = 1 & \ quad, \, i ^ {25} = i ^ {24} \ cdot i = i \
i ^ {79} = i ^ {76} \ cdot i ^ 3 = -i & \ quad, \, i ^ {568} = 1 & \ quad, \, i ^ {946} = i ^ {944} \ cdot i ^ 2 = -1 & \ quad, \, i ^ {- 7} = \ frac {1} {i ^ 7} = \ frac {i} {i ^ 8} = i
\ end {array}
\ end {align *}

A number is divisible by 4 if its last two digits result in a number divisible by 4.

Combined example

Determine \$ \ textrm {Re {z}}, \ textrm {Im {z}}, \ textrm {abs {z}}, \ overline {z} \ textrm {and} \ textrm {arg (z)} \$
\ begin {align *}
z & = 2 \ sqrt [i-1] {i ^ {57}}, \ quad z = 2 \ sqrt [i-1] {i ^ {56} \ cdot i} = 2 \ sqrt [i-1] { i} = 2i ^ {\ frac {1} {i-1}} {=} 2 \ exp ^ {\ frac {1} {i-1} \ ln (i)} \ \
& = 2 \ text {exp} \ left \ {\ frac {-1-i} {(i-1) (- i-1)} \ ln (\ exp ^ {\ frac {\ pi} {2} i }) \ right \} = 2 \ text {exp} \ left \ {- \ frac {1 + i} {2} \ cdot \ frac {\ pi} {2} i \ right \} = 2 \ text {exp } \ left \ {\ frac {\ pi} {4} - \ frac {\ pi} {4} i \ right \} \ \
& = 2 \ exp ^ {\ frac {\ pi} {4}} \ left (\ cos \ Bigl (- \ frac {\ pi} {4} \ Bigr) + i \ sin \ Bigl (- \ frac {\ pi} {4} \ Bigr) \ right) {=} 2 \ exp ^ {\ frac {\ pi} {4}} \ left (\ cos \ Bigl (\ frac {\ pi} {4} \ Bigr) - i \ sin \ Bigl (\ frac {\ pi} {4} \ Bigr) \ right) \ \
& = 2 \ exp ^ {\ frac {\ pi} {4}} \ left (\ frac {1} {\ sqrt {2}} - i \ cdot \ frac {1} {\ sqrt {2}} \ right ) = \ sqrt {2} \ exp ^ {\ frac {\ pi} {4}} - \ sqrt {2} \ exp ^ {\ frac {\ pi} {4}} \ cdot i \ \
\ textrm {Re {z}} & = \ sqrt {2} \ exp ^ {\ frac {\ pi} {4}}, \ quad \ \ textrm {Im {z}} = - \ sqrt {2} \ exp ^ {\ frac {\ pi} {4}}, \ quad \ = 2 \ exp ^ {\ frac {\ pi} {4}}, \ quad \ \ overline {z} = \ sqrt {2} \ exp ^ {\ frac {\ pi} {4}} + \ sqrt {2} \ exp ^ {\ frac {\ pi} {4}} \ cdot i \ \
\ arg (z) & = - \ frac {\ pi} {4}
\ end {align *}

### Complex sequences and series

We want to briefly touch on the subject of "Convergence behavior of complex sequences and series". A complex sequence or series is convergent when its real and imaginary parts converge.

The limit value calculation and the detection of convergence or divergence of sequences and series works with complex numbers in the same way as with real numbers. However, the application of the convergence criteria is limited. There is no comparison criterion for sequences and no major, minor and Leibniz criterion for series and, in general, no upward or downward estimates, since do not exist for complex numbers.

\ begin {align *}
a_n = \ frac {n + 3i} {2n-i} \, \ quad \ lim \ limits_ {n \ to \ infty} {a_n} = \ lim \ limits_ {n \ to \ infty} {\ frac {n + 3i} {2n-i}} = \ lim \ limits_ {n \ to \ infty} {\ frac {n \ left (1+ \ frac {3i} {n} \ right)} {n \ left (2- \ frac {i} {n} \ right)}} = \ frac {1} {2}
\ end {align *}

\ begin {align *}
b_n = \ frac {n + 3i} {2n-ni} \, \ quad \ lim \ limits_ {n \ to \ infty} {b_n} = \ lim \ limits_ {n \ to \ infty} {\ frac {n \ left (1+ \ frac {3i} {n} \ right)} {n \ left (2-i \ right)}} = \ frac {1} {2-i} = \ frac {1 (2 + i) } {4 + 1} = \ frac {2} {5} + \ frac {1} {5} i
\ end {align *}

\ begin {align *}
c_n = \ left (\ frac {3} {4} \ cdot \ text {e} ^ {i \ pi} \ right) ^ n \, \ quad \ lim \ limits_ {n \ to \ infty} {c_n} = \ lim \ limits_ {n \ to \ infty} {\ left (\ frac {3} {4} \ cdot \ text {e} ^ {i \ pi} \ right) ^ n} = \ lim \ limits_ {n \ to \ infty} {\ left (\ frac {3} {4} \ right) ^ n \ cdot \ text {e} ^ {i \ pi n}} = 0
\ end {align *}

\ begin {align *}
\ sum_ {k = 0} ^ {\ infty} & {\ frac {(2 + i) ^ k} {k!}} \, \ quad \ left | \ frac {a_ {k + 1}} {a_k} \ right | = \ left | \ frac {\ frac {(2 + i) ^ {k + 1}} {(k + 1)!}} {\ frac {(2 + i) ^ k} {k!} } \ right | = \ left | \ frac {(2 + i) (2 + i) ^ k \ cdot k!} {k! (k + 1) \ cdot (2 + i) ^ k} \ right | = \ left | \ frac {2 + i} {k + 1} \ right | \ lim \ limits_ {k \ to \ infty} 0 <1 \
& \ Rightarrow \ sum_ {k = 0} ^ {\ infty} {\ frac {(2 + i) ^ k} {k!}} \ Text {converges (absolute).}
\ end {align *}