Complex numbers (en)

Rehersal, complex numbers ...

en pdf complex_en.pdf

en Basic properties of complex numbers Definitions A common, real number is usually illustrated as a point on the so-called number line. The magnitude is represented by the distance from the point in question to zero.

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A complex number z consists of two components. It can be written as a + jb. Here, a and b are real numbers. j is the square root of -1 and is called the imaginary unit. a is the complex number real part Re (Z). b is the imaginary part, Im (Z).

Every complex number can be represented as a point in a two-dimensional coordinate system, the complex plane.

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Number z is represented by a point with coordinates a and b.

The distance from the point to the origin represents the amount or number value |z|.

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or

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The angle α is called the argument of z, arg(z) and as seen in the figure

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or

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We can also express z in polar form, eg with |z| and α. As seen in the figure

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One can then imagine that it's the connecting line between the point and the origin that represents the number. We can see this as a pointer (vector) with the length |z| och en direction that is defined by the angle α.

Basic properties Complex numbers can be treated algebraically, the following rules apply.

Addition komplexadd.jpg

z1z2add.jpg

Figuren visar vad additionen innebär i det kThe figure shows what the addition means in the complexatal planet. Visaren för z blir lika med denThe pointer of z equals the geometriskac summan av visarna för of the pointers of z1 ochand z2. Föor |z| ochand arg(z) gäller de tidigare angivna generella uttryckenapplies the previously mentioned general terms.

Subtraction z1z2sub.jpg

I talplanet blir visaren för z lika med den geometriska skillnaden mellan visarna för In the complex plane z equals the geometric difference between the z1 och z2.

Multiplication MThe multiplikcationsregeln demonstrerar vi enklast med ett rule, we demonstrate most easily with an exeampele.

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MThe multiplikcationen kan också genomföras med talen uttryckta can also be implemented with the numbers expressed in poläar form.

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Detta innebär atThis means that

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Division Algebraiskt genomförs divisionen så här:¶ div.jpg¶ Nu vill man ofta ha resultatet ic the division is implemented like this:¶

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Now, one often wants to have the results in the formen a+jb ochand if så fall förlänger man med nämnarens ko, one extends the denominator with the conjugatkves quantitety a2 - jb2. Då får manThen one gets

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Uttrycks talenIf numbers are expressed in poläar form kommer, the divisionsregeln att se ut så här rule that look like this:

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Några minnesregSome memory rulers
* OmIf z = z1 + z2, så är i allmänheto is generally |z| ≠ |z1| + |z2|( endast omonly if arg(z1) = arg(z2) ärthen |z| = |z1| + |z2| )

* När man skall bilda beloppet av enWhen calculating the amount of a produkct eller en kvot mellan två komplexa tal z1 ochz2 är det i allmänhet onödigt attor a quotient of two complex numbers z1 ochz2 it is generally unnecessary to föirst ta fram det kcalculate the complexa resultatet och sedan bilda beloppet av detta. Man beräknar and then form the amount. One calculates inställetead |z1| ochand |z2| varseparately, föor sig, ty som vi sett gälleras we have seen applies

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Exeampele Exeampele 1 Gör om uttrycket 2 + Redo the expression 2+3/j tillo formen a+jb.

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Exeampele 2 Skriv uttrycketWrite expression z = 6 + jA + 1/(jB) i allmänn the general form för kof complexa tal, samt teckna beloppet av uttrycke numbers, and write a expression for the amount.

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Exeampele 3 BestämDetermine |z| ochand arg(z) omwhen z = z1·z2 ochand z1 = j ochand z2 = -1 -j

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Algebraisktc ex3.jpg

Polärtar ex3polar.jpg

Exeampele 4 z1 = 3 + j5, z2 = 5 + j7. BeräknaCalculate

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Om manIf instället heade multiplicerat med ked with conjugatkve quantiteten hade man fått¶ ex4konjug.jpg¶ Om man jämför med ovanstående ser man aty the calculations had been¶

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If one compares the above one can see that kcomplexk conjugeringen medför mycket mera arbete!¶ ¶ Övningsuppgifter Fråga 1 Åt vilket håll pekar visaren z = -2 + j2 ?¶ rutat.jpg¶ [ Answers and solutions ]¶ Fråga 2 Vad är summan av z1 och z2 om z1 = 1 + j2 och z2 = 2 - j ?¶ [ Answers and solutions ]¶ Fråga 3 Hur lång är visaren 3 + j4 ?¶ [ Answers and solutions ]¶ Fråga 4 Rita visaren z = z1 - z2 om z1 = 1 + j och z2 = 2 + j ?¶ rutat.jpg¶ [ Answers and solutions ]¶ Fråga 5 Hur stor blir Im(z) om z = z1 + z2 ?z1 = 3(1+j) och z2 = 2(1-j) .¶ [ Answers and solutions ]¶ Fråga 6 Hur stor blir |z| om z = z1·z2 ?z1 = 2 + j och z2 = -(2 + j) .¶ [ Answers and solutions ]¶ Fråga 7 Vad blir |3+j4|· |j2| ?¶ [ Svar och lösningar ]¶ Fråga 8 Bestäm |z| och arg(z) om z = z1·z2 och z1 = 1 + j och z2 = -1 + j .¶ rutat.jpg¶ [ Answers and solutions ]¶ Fråga 9 Vad blir z = z1·z2 om z1 = j ochation involves much more work!¶

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Exercises Question 1 In which direction points the complex pointer z = -2 + j2 ?¶

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[ Answers and solutions ]¶

Question 2 What is the sum of z1 and z2 if z1 = 1 + j2 and z2 = 2 - j ?¶

[ Answers and solutions ]¶

Question 3 How long is the pointer 3 + j4 ?¶

[ Answers and solutions ]¶

Question 4 Draw the pointer z = z1 - z2 if z1 = 1 + j and z2 = 2 + j ?¶

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[ Answers and solutions ]¶

Question 5 How large is Im(z) if z = z1 + z2 ?z1 = 3(1+j) and z2 = 2(1-j) .¶

[ Answers and solutions ]¶

Question 6 How large is |z| if z = z1·z2 ?z1 = 2 + j and z2 = -(2 + j) .¶

[ Answers and solutions ]¶

Question 7 What will be |3+j4|· |j2| ?¶

[ Answers and solutions ]¶

Question 8 Determine |z| and arg(z) if z = z1·z2 and z1 = 1 + j and z2 = -1 + j .¶

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[ Answers and solutions ]¶

Question 9 What will be z = z1·z2 if z1 = j and z2 = 1 - j .

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[ Answers and solutions ]

Fråga 10 Vad ärQuestion 10 What is |z| ?

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[ Answers and solutions ]

Fråga 11 Beräkna z.z1 = 2 + j3 ochQuestion 11 Calculate z.z1 = 2 + j3 and z2 = 1 + j .

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[ Answers and solutions ]

This exercise booklet has been given to me by Per-Erik Lindahl. It has been used as an aid in courses of basic circuit theory at KTH.