Multiply Complex Numbers

The complex number consists of imaginary part and real part is defined as the complex number.

The complex number is written the form x + yi where x and y are real numbers and i is defined as the imaginary unit.

The normal numbers in the complex numbers is extended by using extra numbers.

The complex numbers are used in quantum physics, engineering and applied mathematics.

Multiply Complex Number:

Multiplication: (a + bi)(c + di) = (ac - bd) + (bc + ad)i

Examples on Multiply Complex Numbers:

Example 1:

Solve (2 + 2i)(5 + 3i)

Solution:

The multiplication of the complex numbers in the arithmetic form is given as,

(2 + 2i)(5 + 3i) = 10 + 6i + 10i + 6i^2

= 10 + 16i - 6   { Since i^2  = -1 }

= 4 + 16i.

The complex number multiplication answer is 4 + 16i.

Example 2:

Solve

(3 + 7i)(4 + 3i)

Multiplication of the complex number in algebraic form is given as,

(3 + 7i)(4 + 3i) = 12 + 9i + 28i + 21i^2

= 12 + 37i - 21

= -9 + 37i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+37i.

Example 3:

Solve

(4 + 7i)(3 + 3i)

Multiplication of the complex number in algebraic form is given as,

(4 + 7i)(3 + 3i) = 12 + 21i + 12i + 21i^2

= 12 + 33i - 21

= -9 + 33i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+33i.

Example 4:

Solve

(4 + 3i)(3 + 7i)

Multiplication of the complex number in algebraic form is given as,

(4 + 3i)(3 + 7i) = 12 + 9i + 28i + 21i^2

= 12 + 37i - 21

= -9 + 37i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+37i.

Example 5:

Solve

(2 + 5i)(4 + 6i)

Multiplication of the complex number in algebraic form is given as,

(2 + 5i)(4 + 6i) = 8 + 20i + 12i + 30i^2

= 8 + 32i - 30

= -28 + 32i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -28+32i.

Practice Algebra Calculus

Just as algebra introduces students to new ways of thinking about arithmetic problems (by way of variables, equations, functions and graphs), calculus may be seen as introducing new ways of thinking about algebra problems.

A branch of mathematics, algebra calculus  mainly deals with derivatives, integrals, limits, infinite series, differential equations, and functions, is called calculus. It is used for computing rate of change of velocity and rate of change of acceleration, the slope of a curve, and optimization

Types of Calculus and Problems in Algebra Problems:


Calculus can be classified into two major types as follows

Differential calculus
Integral calculus

Ex 1: Compute the values of Δy and dy if y = f(x) = x^3 + x^2 − 2x + 1  where x changes (i) from 2 to 2.05 and (ii) from 2 to 2.01

Sol :    (i) We have f(2) = 2^3 + 2^2 − 2(2) + 1 = 9

f(2.05) = (2.05)^3 + (2.05)^2 − 2(2.05) + 1 = 9.717625.

And Δy = f(2.05) − f(2) = 0.717625.

In general dy = f ′(x) dx = (3x^2 + 2x − 2)dx

When x = 2, dx = Δx = 0.05 and dy = [(3(2)^2+2(2)−2] 0.05 = 0.7

(ii) f(2.01) = (2.01)^3 − (2.01)^2 − 2(2.01) + 1 = 9.140701

Δy = f(2.01) − f(2) = 0.140701

When dx = Δx = 0.01, dy = [3(2)^2 + 2(2) − 2]0.01 = 0.14

Remark :   The approximation Δy ≈ dy becomes better as Δx becomes smaller . Also dy was easier than to compute Δy. For more complicatedfunctions it may be impossible to compute Δy exactly. In such cases theapproximation by calculus is especially useful.
Example 2 Algebra Calculus Problems:

Ex 2 :  Use differentials to find an approximate value for

`root(3)65` .

Let y = f(x) =`3sqrtx`=x.^1/3 Then dy =1/3x ^-2/3 dx

Since f(64) = 4. We take x = 64 and dx = Δx = 1

This gives dy =1/3 (64) ^−2/3 (1)

=1/3(16)       = 1/48

365      = f(64 + 1) ≈ f(64) + dy = 4 +1/48

≈ 4.021

Practice Problems

Pro 1:    If f(x) = x^4, then find f'(x)

Sol :              f(x) = x^4

f'(x) = 4x^3

Pro 2:    If f(x) = 5x3^ + 2x^2, then find f'(x).

Sol :              f(x) = 5x^3 + 2x^2

f'(x) = 15x^2 + 4x

Pro 3:   x7  dx  =  (x^7+1) / (7 + 1) + c

Sol :                   = x^8/8 + c

Algebra with Negative Numbers

The algebra is a division of mathematics, which can be explained as calculation of arithmetic operations. It is also referred to that character type of abstract algebra pattern. The negative number is indicating the symbol "-". For example the number -6, is the negative number. The elementary algebra is the division of the algebra.  In the subsequent we see in detail about algebra with negative numbers.

Example Problem for Algebra with Negative Number

Example 1:

Given two number -23x and -43x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -23x + (-43x)

= - (23x+43x)

= - (66x)

Answer for this problem is -66x.

Example 2:

Given two number -120x and -56x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -102x - (-56x)

= -102x+56x

= - 46x

Answer for this problem is -46x.


More over Problem for Algebra with Negative Number

Example 3:

Given two number -430x and -123x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -430x + (-123x)

= - (430x+123x)

= - (553x)

Answer for this problem is -553x.

Example 4:

Given two number -1230x and -1405x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -1230x - (-1405x)

= - 1230x + 1405x

= 175x

Answer for this problem is 175x.

Example 5:

Given two number -2345x and -450x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -2345x + (-450x)

= - (2345x+450x)

= - (2795x)

Answer for this problem is -2795x.

Example 6:

Given two number -5623x and -2345x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -5623x - (-2345x)

= - 5623x + 2345x

= - 3278x

Answer for this problem is -3278x.

Multiplication Problems

Introduction to multiplication problems:
Multiplication – It is a repeated addition.
For example 2 * 3 = 6 this is nothing but 2+2+2 = 6 or 3+ 3 =6
The order of multiplication problems does not affect the answer.The answer will not be changed even though the performing order changed.
Distribution in multiplication problems 
            To multiply 4(10+5), we multiply 4*10 and 4*5, adding the result. Thus,
                             4(10+5) = (4*10) + (4*5)
                                           =40+20
                                           =60
This is nothing but the distributive property of multiplication.
It will be easy to do multiplication if we are well known with multiplication tables.
When we multiply two numbers, there is no matter which number is first or second but the answer will be same.

More on Multiplication Problems

Example 1:

Multiply 46 * 3
Solution:
First multiply the ones digit with 3 and next tens digit with 3
If we have any remainder carry to next digit and add with it
46 * 3 =138

Example 2:

Multiply 45 * 5
Solution:
First multiply the ones digit with 5 and next tens digit with 5
If we have any remainder carry to next digit and add with it
45 * 5 = 225

Example 3:

Multiply  100 * 39
Solution:
This multiplication problem is very simple
Just multiply 1*39 and add two zero at the end
So 3900 is the answer

Example 4:

Multiply  100 * 65
Solution:
This multiplication problem is very simple
Just multiply 1*65 and add two zero at the end
So 6500 is the answer.

Example 5:

Multiply:   66 1* 23
Solution:
      6 6 1
(*)    2 3
-------------
     1983
     1322
-------------
  1 5 203
------------

Example 6:

Multiply:  2546 * 121
Solution:
                      2 5 4 6
              (*)       1 2 1
          -------------------------
                  2 5 4 6
               5 0 9 2
           2 5 4 6
--------------------------------
          1 0 1 8 4
----------------------------------

Example 7:

Multiply 45 * 9
Solution:
First multiply the ones digit with 9 and next tens digit with 9
If we have any remainder carry to next digit and add with it
45 * 9 = 405

Example 8:

Multiply  100 * 65
Solution:
This multiplication problem is very simple
Just multiply 1*65 and add two zero at the end
So 6500 is the answer.

Example 9:

Multiply 83 * 6
Solution:
First multiply the ones digit with 6 and next tens digit with 6
If we have any remainder carry to next digit and add with it
83 * 6 = 498

Example 10:

Multiply  1000 * 869
Solution:
This multiplication problem is very simple
Just multiply 1*869 and add three zero at the end
So 869000 is the answer.

Equations with Exponent

Introduction to solving equations with exponents

Exponent equations are the equations in which variable appear as an exponent.
To solve these equations rules and laws of exponents are used. Exponent equations are of two types
(1) Exponent equations in which bases are same
(2) Exponent equations in which bases are different.

Steps to Solve Equations with Exponent

Solving Exponential Equations of the same base

1) Ignore the bases, and simply set the exponents equal to each other

2) Solve for the variable

 When the bases of the terms are different

1) Ignore the exponents; rewrite both of the bases as powers of same number. For    example if there are 2 and 4 in the bases, then convert base 4, in to base 2 by writing it again as (2)^2

2) once the bases are same , ignore them

3) Equalize the exponents

4) Solve for variable

Simple Problems of Equations with Exponents

Solve for variable Answer 1. 3m  =  35 Since the bases are the same, set the exponents equal to one another: m = 5 2. 5t   = 125 125can be expressed as a power of 5: 5t  = 53 t = 3  3.  493y=343 49 and 343 can be expressed as a power of 7: [(7)2]3y = 73 76y = 73 6y = 3 y = 1/2

More Problems of Equations with Exponents

Solve for x. Answer 1.  52x+1  =  53x-2 Since the bases are the same, set the exponents equal to one another: 2x + 1 = 3x - 2 3 = x 2.  32x-1  = 27x  27 can be expressed as a power of 3: 32x-1  = 33x 2x - 1 = 3x -1 = x 3.   43x-8  = 162x 16 can be expressed as a power of 4: 43x-8= [(4)2]2x 3x - 8 = 4x   -8 = x

Practicing Integers

Introduction to Practicing Integers:
                 An integer is a set of whole numbers. Whole numbers above zero is said to be positive integers denoted as ‘+’ sign and whole numbers below zero is said to be negative integers denoted as ‘-‘ in a number line. Zero is said to be neither negative nor positive integer and it does not constant. Here, integers can be performed with four basic operations such as addition, subtraction, multiplication, and division. The positive integers can be written with or without the sign. Let us see practicing integers in this article.

Practicing Integer Problems - Practicing Adding Integers in Math

Adding same signed Integers:

Example 1:

4 + 2

Solution:

The absolute value of 4 and 2 is 4 and 2. Put the positive sign before the sum of two integers.

4 + 2 = 6

Therefore, the solution for adding 4 and 2 is 6.

Example 2:

(-4) + (-8)

Solution:

The absolute value of -4 and -8 is 4 and 8. Put the negative sign before the sum of two integers.

(-4) + (-8) = - (4 + 8) = - 12

Therefore, the solution for adding -4 and  -8 is -12.

Adding different signed Integers:

Example 3:

5 + (-5)

Solution:

The absolute value of 5 and -5 is 5 and 5. Put the sign of larger number sign before the sum of two integers.

5 – 5 = 0

Therefore, the solution for adding 5 + (-5) is 0.

Practicing Subtracting Integers in Math

Example 4:

7 - (-2)

Solution:

The absolute value of 7 and -2 is 7 and 2. Subtract the integers and put sign of larger integer before the sum.

7 – (-2) = 7 + 2 = 9

Therefore, the solution for subtracting the above integers are 9.

Practicing Multiplying Integers in Math

Multiplying same signed Integers

Example 5:

4 × 3

Solution:

The absolute value of 4 and 3 is 4 and 3. Place the same sign before the answer as it is in the given problem.

4 × 3 = 12

Therefore, the oslution for multiplying 4 and 3 is 12.

Example 6:

(-5) × (-6)

Solution:

The absolute value of -5 and -6 is 5 and 6.

- 5 × - 6 = - 30

Place the same sign before the answer as it is in the given problem -30.

Multiplying different signed Integers:

Example 7:

(-5) × (4)

Solution:

The absolute value of -5 and 4 is 5 and 4.

-5 × 4 = - 20

There is a negative sign before one of the integers in a given problem so put the negative sign before the product.

Therefore, the solution for multiplying -5 and 4 is -20.

Practicing Dividing Integers in Math

Dividing same signed Integers:

Example 8:

16 ÷ 4

Solution:

The absolute value of 16 and 4 is same.

The integer 16 is the multiple of 4 and 4.

Therefore, the solution for dividing 16 ÷ 4 is 4.

Dividing different signed Integers:

Example 9:

28 ÷ -7

Solution:

The absolute value of 28 and -7 is 28 and 7. There is a negative sign before the integers in a given problem so put the negative sign before the quotient.

The integer 28 is the multiple of 4 and 7.

Therefore, the solution for dividing 28 ÷ -7 is -4.

Problems to Practicing Integers

1. Add the two integers 3 + 5

Key: 8

2. Subtract the two integers (– 5) – (- 4)

Key: -1

3. Multiply the two integers 7 × -3

Key: -21

4. Divide the integer 15 by -5

Key: - 3