Instruction

1

If the module is presented in the form of a continuous function, then the value of its argument can be both positive and negative: |x| = x, x ≥ 0; |x| = - x, x < 0. Consequently, the number in brackets, accepts any character.

2

Module zero is zero, and the module is any positive number – to him. If the argument is negative, then after opening the brackets, its sign changes from minus to plus. Based on this, the conclusion follows that the modules of opposite integers is equal to: |x| = |x| = x.

3

The module of a complex number is given by: |a| = √b 2 + c 2, and |a + b| ≤ |a| + |b|. If the argument is present in multiples of a positive integer, it can be taken out of the sign brackets, for example: |4*b| = 4*|b|.

4

The negative module can not be, therefore, any negative number is converted to positive: |x| = x, |-2| = 2, |-1/7| = 1/7, |-2,5| = 2,5.

5

If the argument is in the form of complex numbers, for computational convenience it is allowed to change the order of the member expressions, enclosed in square brackets: |2-3| = |3-2| = 3-2 = 1, since (2-3) is less than zero.

6

Raised to the power of argument at the same time is under the root sign are of the same order – it is solved by using module: √a2 = |a| = ±a.

7

If you face the problem, which does not indicate a condition of disclosure of the brackets module, to get rid of them is not necessary – this will be the end result. But if you want to reveal them, it is necessary to indicate sign ±. For example, you need to find the value of √(2 * (4-b)) 2. His decision as follows: √(2 * (4-b)) 2 = |2 * (4-b)| = 2 * |4-b|. Since the expression of 4-b is unknown, it should be left in brackets. If you add an additional condition, for example, |4-b| > 0, then the end result 2 * |4-b| = 2 *(4 - b). As the unknown element can also be set to a specific number, which should be taken into consideration because it will affect the sign of the expression.

# Advice 2 : How to find the modulus of a complex number

Real numbers will not be enough to solve any quadratic equation. The simplest of quadratic equations with no roots among the real numbers is x^2+1=0. In its decision, it turns out that x=±sqrt(-1), and according to the laws of elementary algebra, extracting the root of an even degree of a negative

**number**is impossible.You will need

- paper;
- - handle.

Instruction

1

In this case, there are two ways: the first is to follow the established restrictions and assume that this equation has no roots; the second is to extend the system of real numbers to such an extent that the equation will have a root.So, the notion of complex numbers of the form z=a+ib, in which (i^2)=-1, where i is the imaginary unit. The numbers a and b are called, respectively, the real and imaginary parts

**of the number**z Rez and Imz.An important role in the actions with complex**numbers**mi play a**number of**complex-conjugate. Conjugate to the complex number z=a+ib is called zs=a-ib, i.e. the number with the opposite sign before the imaginary unit. So, if z=3+2i, z=3-2i.Any real number is a special case of complex**numbers**, imaginary part is zero. 0+i0 is a complex number equal to zero.2

Complex

**numbers**can be added and multiplied just as is done with algebraic expressions. Thus the usual laws of addition and multiplication remain valid. Let z1=a1+ib1, z2=a2+ib2.1. Addition and subtraction.z1+z2=(a1+a2)+i(b1+b2) z1-z2=(a1-a2)+i(b1-b2). 2. Multiplication.z1*z2=(a1+ib1)(a2+ib2)=a1a2+ia1b2+ia2b1+(i^2)b1b2=(a1a2-b1b2)+i(a1b2+a2b1).When multiplying simply open the brackets and use the definition i^2=-1. The product of complex conjugate numbers is a real number: z*z=(a+ib)(a-ib)==a^2-(i^2)(b^2) = a^2+b^2.3

3. Division.To bring the quotient z1/z2=(a1+ib1)/(a2+ib2) to the standard view you need to get rid of the imaginary unit in the denominator. The easiest way is to multiply the numerator and the denominator by the number, conjugate denominator: ((a1+ib1)(a2-ib2))/((a2+ib2)(a2-ib2))=((a1a2+b1b2)+i(a2b1-a1b2))/(a^2+b^2)= =(a1a2+b1b2)/(a^2+b^2)+i(a2b1-a1b2)/(a^2+b^2).Operations of addition and subtraction and multiplication and division are inverses.

4

*Example. Calculate (1-3i)(4+i)/(2-2i)=(4-12i+i+3)(2+2i)/((2-2i)(2+2i))=(7-11i)(2+2i)/(4+4)=(14+22)/8+i(-22+14)/8=9/2-rassmotrite geometric interpretation of complex numbers. On the plane with a Cartesian coordinate system 0xy each complex number z=a+ib, you must assign a point of the plane with coordinates a and b (see Fig. 1). The plane, which implemented such a correspondence, called the complex plane. On the axis 0x located a valid*

**number**, so it is called a valid axis. On the axis 0y is located imaginary

**number**, it is called the imaginary axis.

5

C in each point z of the complex plane is associated the radius-vector of this point. The length of the radius vector representing the complex number z, called modeling=|z| of a complex

**number**; the angle between the positive direction of the real axis and the direction of the vector 0Z is called the argument argz of this complex**number**.6

The argument is a complex

**number**is positive if it is measured from the positive direction of the 0x axis in a counterclockwise direction, and negative in the opposite direction. One complex number corresponds to many values of the argument argz+2pcs. These values are considered to be the principal value argz lying in the range from –p to p. Conjugate complex**numbers**z and z have equal modules, and their arguments are equal in absolute value but different sign.7

Thus, |z|^2=a^2+b^2, |z|=sqrt(a^2+b^2). So, if z=3-5i, then |z|=sqrt(9+25)=6. In addition, since z*z=|z|^2=a^2+b^2, it becomes possible to calculate the modules as much as comprehensive terms in which the imaginary unit may appear multiple times.Since z=(1-3i)(4+i)/(2-2i)=9/2-i, a direct computation module z gives |z|^2=81/4+1=85/4 and |z|=sqrt(85)/2.Bypassing the stage of computing the expression, given that zs=(1+3i)(4-i)/(2+2i), we can write |z|^2=z*zs==(1-3i)(1+3i)(4+i)(4-i)/((2-2i)(2+2i))=(1+9)(16+1)/(4+4)=85/4 and |z|=sqrt(85)/2.