وحدة تخيلية

في الرياضيات، الوحدة التخيلية Imaginary unit، هي التي تتيح توسيع مجموعة الأعداد الحقيقية إلى مجموعة الأعداد المركبة، والتي تمكن من إيجاد جذر واحد على الأقل لكثيرات الحدود د(س). يرمز لها عادة بالرمز $i$ . ويمكن تعريفها على أنها القيمة التي تحقق $x^{2}+1=0$ . والمثال البسيط لاستخدام $i$ في عدد مركب هو $2+3i$ .

$i$ في المستوى العقدي أو الديكارتي; تمثل الأعداد الحقيقية على المحور الأفقي، والأعداد التخيلية على المحور العمودي

Imaginary numbers are an important mathematical concept; they extend the real number system $\mathbb {R}$ to the complex number system $\mathbb {C}$ , in which at least one root for every nonconstant polynomial exists (see Algebraic closure and Fundamental theorem of algebra). Here, the term "imaginary" is used because there is no real number having a negative square.

There are two complex square roots of −1: $i$ and $-i$ , just as there are two complex square roots of every real number other than zero (which has one double square root).

In contexts in which use of the letter $i$ is ambiguous or problematic, the letter $j$ is sometimes used instead. For example, in electrical engineering and control systems engineering, the imaginary unit is normally denoted by $j$ instead of $i$ , because $i$ is commonly used to denote electric current.^[1]

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التعريف

قوى $i$ تـُنتج قيم دوارة:
$...$ (يتكرر النمط من المنطقة الزرقاء)
$i -3 = i$
$i -2 = -1$
$i -1 = - i$
$i 0 = 1$
$i 1 = i$
$i 2 = -1$
$i 3 = - i$
$i 4 = 1$
$i 5 = i$
$i 6 = -1$
$...$ (يتكرر النمط من المنطقة الزرقاء)

The imaginary number $i$ is defined solely by the property that its square is −1:

i^{2}=-1.

With $i$ defined this way, it follows directly from algebra that $i$ and $-i$ are both square roots of −1.

Although the construction is called "imaginary", and although the concept of an imaginary number may be intuitively more difficult to grasp than that of a real number, the construction is valid from a mathematical standpoint. Real number operations can be extended to imaginary and complex numbers, by treating $i$ as an unknown quantity while manipulating an expression (and using the definition to replace any occurrence of $i^{2}$ with −1). Higher integral powers of $i$ can also be replaced with $-i$ , 1, $i$ , or −1:

i^{3}=i^{2}i=(-1)i=-i

i^{4}=i^{3}i=(-i)i=-(i^{2})=-(-1)=1

or, equivalently,

i^{4}=(i^{2})(i^{2})=(-1)(-1)=1

i^{5}=i^{4}i=(1)i=i

Similarly, as with any non-zero real number:

i^{0}=i^{1-1}=i^{1}i^{-1}=i^{1}{\frac {1}{i}}=i{\frac {1}{i}}={\frac {i}{i}}=1

As a complex number, $i$ can be represented in rectangular form as $0+1i$ , with a zero real component and a unit imaginary component. In polar form, $i$ can be represented as $1\times e^{i\pi /2}$ (or just $e^{i\pi /2}$ ), with an absolute value (or magnitude) of 1 and an argument (or angle) of ${\tfrac {\pi }{2}}$ radians. (Adding any multiple of $2 π$ to this angle works as well.) In the complex plane (also known as the Argand plane), which is a special interpretation of a Cartesian plane, $i$ is the point located one unit from the origin along the imaginary axis (which is orthogonal to the real axis).

$i$ و $- i$

Being a quadratic polynomial with no multiple root, the defining equation $x^{2}=-1$ has two distinct solutions, which are equally valid and which happen to be additive and multiplicative inverses of each other. Once a solution $i$ of the equation has been fixed, the value $-i$ , which is distinct from $i$ , is also a solution. Since the equation is the only definition of $i$ , it appears that the definition is ambiguous (more precisely, not well-defined). However, no ambiguity will result as long as one or other of the solutions is chosen and labelled as " $i$ ", with the other one then being labelled as $-i$ .^[2] After all, although $-i$ and $+i$ are not quantitatively equivalent (they are negatives of each other), there is no algebraic difference between $+i$ and $-i$ , as both imaginary numbers have equal claim to being the number whose square is −1.

In fact, if all mathematical textbooks and published literature referring to imaginary or complex numbers were to be rewritten with $-i$ replacing every occurrence of $+i$ (and, therefore, every occurrence of $-i$ replaced by $-(-i)=+i$ ), all facts and theorems would remain valid. The distinction between the two roots $x$ of $x^{2}+1=0$ , with one of them labelled with a minus sign, is purely a notational relic; neither root can be said to be more primary or fundamental than the other, and neither of them is "positive" or "negative".^[3]

The issue can be a subtle one. One way of articulating the situation is that although the complex field is unique (as an extension of the real numbers) up to isomorphism, it is not unique up to a unique isomorphism. Indeed, there are two field automorphisms of $C$ that keep each real number fixed, namely the identity and complex conjugation. For more on this general phenomenon, see Galois group.

المصفوفات

Some imaginary units correspond to points

(x, y)

on the hyperbola

xy = -1

.

Using the concepts of matrices and matrix multiplication, imaginary units can be represented in linear algebra. The value of 1 is represented by an identity matrix $I$ and the value of $i$ is represented by any matrix $J$ satisfying $J 2 = - I$ . A typical choice is

I={\begin{pmatrix}1&0\\0&1\end{pmatrix}},\qquad J={\begin{pmatrix}0&-1\\1&0\end{pmatrix}}\,.

More generally, a real-valued

2 \times 2

matrix

J

satisfies

J 2 = - I

if and only if

J

has a matrix trace of zero and a matrix determinant of one, so

J

can be chosen to be

J={\begin{pmatrix}z&x\\y&-z\end{pmatrix}}\,,

whenever

- z 2 - xy = 1

. The product

xy

is negative because

xy = -(1 + z 2)

; thus, the points

(x, y)

lie on hyperbolas determined by

z

in quadrant II or IV.

Matrices larger than $2 \times 2$ can be used. For example, $I$ could be chosen to be the $4 \times 4$ identity matrix with $J$ chosen to be any of the three $4 \times 4$ Dirac matrices for spatial dimensions, $γ 1, γ 2, γ 3$ .

Regardless of the choice of representation, the usual rules of complex number mathematics work with these matrices because $I \times I = I$ , $I \times J = J$ , $J \times I = J$ , and $J \times J = - I$ . For example,

{\begin{aligned}J^{-1}&=-J\,,\\\left(aI+bJ\right)+\left(cI+dJ\right)&=(a+c)I+(b+d)J\,,\\\left(aI+bJ\right)\times \left(cI+dJ\right)&=(ac-bd)I+(ad+bc)J\,.\end{aligned}}

الاستخدام الصحيح

The imaginary unit is sometimes written ${\sqrt {-1}}$ in advanced mathematics contexts^[2] (as well as in less advanced popular texts). However, great care needs to be taken when manipulating formulas involving radicals. The radical sign notation is reserved either for the principal square root function, which is only defined for real $x\geq 0$ , or for the principal branch of the complex square root function. Attempting to apply the calculation rules of the principal (real) square root function to manipulate the principal branch of the complex square root function can produce false results:^[4]

-1=i\cdot i={\sqrt {-1}}\cdot {\sqrt {-1}}={\sqrt {(-1)\cdot (-1)}}={\sqrt {1}}=1\qquad {\text{(incorrect).}}

Similarly:

{\frac {1}{i}}={\frac {\sqrt {1}}{\sqrt {-1}}}={\sqrt {\frac {1}{-1}}}={\sqrt {\frac {-1}{1}}}={\sqrt {-1}}=i\qquad {\text{(incorrect).}}

Generally, the calculation rules

{\sqrt {a}}\cdot {\sqrt {b}}={\sqrt {a\cdot b}}

and

{\frac {\sqrt {a}}{\sqrt {b}}}={\sqrt {\frac {a}{b}}}

are guaranteed to be valid for real, positive values of $a$ and $b$ only.^[5]^[6]^[7]

When $a$ or $b$ is real but negative, these problems can be avoided by writing and manipulating expressions like $i{\sqrt {7}}$ , rather than ${\sqrt {-7}}$ . For a more thorough discussion, see square root and branch point.

الخصائص

الجذور التربيعية

The two square roots of

i

in the complex plane

The three cube roots of

i

in the complex plane

Just like all nonzero complex numbers, $i$ has two square roots: they are^[أ]

\pm \left({\frac {\sqrt {2}}{2}}+{\frac {\sqrt {2}}{2}}i\right)=\pm {\frac {\sqrt {2}}{2}}(1+i).

Indeed, squaring both expressions yields:

{\begin{aligned}\left(\pm {\frac {\sqrt {2}}{2}}(1+i)\right)^{2}\ &=\left(\pm {\frac {\sqrt {2}}{2}}\right)^{2}(1+i)^{2}\ \\&={\frac {1}{2}}(1+2i+i^{2})\\&={\frac {1}{2}}(1+2i-1)\ \\&=i.\end{aligned}}

Using the radical sign for the principal square root, we get:

{\sqrt {i}}={\frac {\sqrt {2}}{2}}(1+i).

الجذور التكعيبية

The three cube roots of $i$ are:^[9]

-i,

{\frac {\sqrt {3}}{2}}+{\frac {i}{2}},

and

-{\frac {\sqrt {3}}{2}}+{\frac {i}{2}}.

Similar to all the roots of 1, all the roots of $i$ are the vertices of regular polygons, which are inscribed within the unit circle in the complex plane.

الضرب والقسمة

Multiplying a complex number by $i$ gives:

i(a+bi)=ai+bi^{2}=-b+ai.

(This is equivalent to a 90° counter-clockwise rotation of a vector about the origin in the complex plane.)

Dividing by $i$ is equivalent to multiplying by the reciprocal of $i$ :

{\frac {1}{i}}={\frac {1}{i}}\cdot {\frac {i}{i}}={\frac {i}{i^{2}}}={\frac {i}{-1}}=-i~.

Using this identity to generalize division by $i$ to all complex numbers gives:

{\frac {a+bi}{i}}=-i(a+bi)=-ai-bi^{2}=b-ai.

(This is equivalent to a 90° clockwise rotation of a vector about the origin in the complex plane.)

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الأسس

The powers of $i$ repeat in a cycle expressible with the following pattern, where $n$ is any integer:

i^{4n}=1

i^{4n+1}=i

i^{4n+2}=-1

i^{4n+3}=-i,

This leads to the conclusion that

i^{n}=i^{(n{\bmod {4}})}

where mod represents the modulo operation. Equivalently:

i^{n}=\cos(n\pi /2)+i\sin(n\pi /2)

Although we do not give the details here, if one chooses branch cuts and principal values to support it then this last equation would apply to all complex values of $n$ .^[10]

العاملي

The factorial of the imaginary unit $i$ is most often given in terms of the gamma function evaluated at $1+i$ :^[11]^[12]^[13]

i!=\Gamma (1+i)=i\Gamma (i)\approx 0.498015668-0.154949828i.

^[14]

The magnitude of this number is

|\Gamma (1+i)|={\sqrt {\frac {\pi }{\sinh \pi }}}=0.521564046\ldots ,

^[15]

while its argument is

\arg {\Gamma (1+i)}=\lim _{n\to \infty }{\biggl (}\ln {n}-\sum _{k=1}^{n}\operatorname {arccot} {k}{\biggr )}\approx -0.301640320.

^[16]

عمليات أخرى

Many mathematical operations that can be carried out with real numbers can also be carried out with $i$ , such as exponentiation, roots, logarithms, and trigonometric functions. The following functions are well-defined, single-valued functions when $x$ is a positive real number.

A number raised to the $ni$ power is:

x^{ni}=\cos(n\ln x)+i\sin(n\ln x).

The $ni th$ root of a number is:

{\sqrt[{ni}]{x}}=\cos \left({\frac {\ln x}{n}}\right)-i\sin \left({\frac {\ln x}{n}}\right)~.

The cosine of $ni$ is:

\cos ni=\cosh n={\frac {1}{2}}\left(e^{n}+{\frac {1}{e^{n}}}\right)={\frac {e^{2n}+1}{2e^{n}}}\,,

which is a real number when $n$ is a real number.

The sine of $ni$ is:

\sin ni=i\sinh n={\frac {1}{2}}\left(e^{n}-{\frac {1}{e^{n}}}\right)i={\frac {e^{2n}-1}{2e^{n}}}i\,,

which is a purely imaginary number when $n$ is a real number.

In contrast, many functions involving $i$ , including those that depend upon $log i$ or the logarithm of another complex number, are complex multi-valued functions, with different values on different branches of the Riemann surface the function is defined on.^[17] For example, if one chooses any branch where $log i = π i /2$ then one can write

\log _{i}x=-{\frac {2i\ln x}{\pi }}\,,

when $x$ is a positive real number. When $x$ is not a positive real number in the above formulas then one must precisely specify the branch to get a single-valued function; see complex logarithm.

التاريخ

Designating square roots of negative numbers as "imaginary" is generally credited to René Descartes, and Isaac Newton used the term as early as 1670.^[18]^[19] The $i$ notation was introduced by Leonhard Euler.^[20]

انظر أيضاً

ملاحظات

^ To find such a number, one can solve the equation ${\textstyle (x+iy)^{2}=i}$ where $x$ and $y$ are real parameters to be determined, or equivalently $x^{2}+2ixy-y^{2}=i.$ Because the real and imaginary parts are always separate, we regroup the terms, $x^{2}-y^{2}+2ixy=0+i.$ By equating coefficients, separating the real part and imaginary part, we get a system of two equations: ${\begin{aligned}x^{2}-y^{2}&=0\\[3mu]2xy&=1.\end{aligned}}$ Substituting $y={\tfrac {1}{2}}x^{-1}$ into the first equation, we get $x^{2}-{\tfrac {1}{4}}x^{-2}=0$ $\implies 4x^{4}=1.$ Because $x$ is a real number, this equation has two real solutions for $x$ : $x={\tfrac {1}{\sqrt {2}}}$ and $x=-{\tfrac {1}{\sqrt {2}}}$ . Substituting either of these results into the equation $2xy=1$ in turn, we will get the corresponding result for $y$ . Thus, the square roots of $i$ are the numbers ${\tfrac {1}{\sqrt {2}}}+{\tfrac {1}{\sqrt {2}}}i$ and $-{\tfrac {1}{\sqrt {2}}}-{\tfrac {1}{\sqrt {2}}}i$ .^[8]