Complex Analysis Problems

Complex analysis is the functions of the complex numbers. Complex numbers have the real and also the imaginary parts. The complex numbers can be represented as x +i y, where x denoted as the real part and y denoted as the imaginary part. The complex numbers comprises the addition of two complex numbers, subtraction of two complex numbers, multiplication of two complex numbers and also the division of two complex numbers. The conception of the complex numbers is the reflection of the fact. This article has the functions of the complex numbers.

Examples for learn complex analysis problems:

Example 1 to learn complex analysis problems:

Compute the value for the complex number f (x) = (75+42i) + (90+72i).

Solution:

The given complex number is f (x) = (75+42i) + (90+72i).

Step 1: (75+42i) + (90+72i) = (75+90) + (42i + 72i)

Step 2: (75+42i) + (90+72i) = 165 + (42i + 72i)

Step 3: (75+42i) + (90+72i) = 165 +114i

The value for the complex number f (x) = (75+42i) + (90+72i) is f (x) = 165 +114i.

Example 2 to learn complex analysis problems:

Resolve the value for the complex numbers f (x) = (13+i) x (11-i).

Solution:

The given complex numbers are f (x) = (13+i) x (11-i).

Step 1: (13+i) x (11-i) = 13(11 - i) + i (11 -i)

Step 2: (13+i) x (11-i) = (143- 13 i) + (11i - i2)

Step 3: (13+i) x (11-i) = 143 - 13i + 11i - (-1) (where i2= -1)

Step 4: (13+i) x (11-i) = (143+1) + (-13i +11i)

Step 5: (13+i) x (11-i) = 144 - 2i

The value for the complex number f (x) = (13+i) x (11-i) is f (x) = 144 - 2i.

Example 3 to learn complex analysis problems:

Calculate the value for the complex number f (x) = (180+53i) - (82+ 7i).

Solution:

The given complex number is f (x) = (180+53i) - (82+ 7i).

Step 1: (180+53i) - (82+ 7i) = (180-82) + (53i - 7i)

Step 2: (180+53i) - (82+ 7i) = 98 +46i

The value for the complex number f (x) = (180+53i) - (82+ 7i) is f (x) = 98 +46i.

Practice problem for learn complex analysis problems:

Compute the value for the complex numbers f (x) = (45+29i)-(2+7i).

Answer: f (x) = 43- 22i

Predict the value for the complex numbers f (x) = (145+88i) + (33+39i).

Answer: f (x) = 178+127i

Compute the value for the complex numbers f (x) = (5+3i) x (2+i).

Answer: f (x) = 7+11i

Calculate Ratio Math

In mathematics, a ratio expresses the magnitude of quantities relative to each other. Specifically, the ratio of two quantities indicates how many times the first quantity is contained in the second and may be expressed algebraically as their quotient.

Example:

For every Spoon of sugar, you need 2 spoons of flour (1:2)

Source: Wikipedia

Definition- calculate ratio math:

The ration of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the terms of the ratio.

Concept - calculate ratio math:

The numeric ratio of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the conditions of the numeric ratio.

Types of ratio- calculate ratio math:

Compounded ratio in math.

Duplicate ratio in math.

Triplicate ratio in math.

Define compounded ratio in math:

Regular format for compounded numeric ratio is

p/q *r/s=(pr)/(qs)

Example for compounded ratio in math:

How to calculate ratios: 7/5*3/3

Solution:

=7/5*3/3

=(7*3)/(5*3)

=21/15

=7/5 or 7:5

Define duplicate ratio in math:

Regular format for duplicate ratio is

(r/s)*(r/s)=(r^2)/(s^2)

Example for duplicate ratio in math:

How to calculate ratios: 6/5*6/5

Solution:

=6/5*6/5

=(6^2)/(5^2)

=36/25 or 36:25

Define triplicate ratio in math:

Regular format for triplicates ratio is

r/sr/sr/s=(r^3)/(s^3)

Example for triplicate ratios in math:

How to calculate ratios: 5/7 * 5/7 * 5/7

Solution:

=(5*5*5)/(7*7*7)

=(5^3)/(7^3) or 5^3:7^3

=125/343   or 125:343

More about ratios- calculate ratio math:

Define Inverse ratio in math:

It is often want to estimate the numbers of a ratio in the inverse order. To perform this, we simply switch over the numerator and the denominator. Therefore, the inverse of 21:7 is 7:21. When the terms of a ratio are switch over are called  the INVERSE NUMERIC RATIO results.

Example problem for inverse ratios in math:

In math, how to calculate inverse numeric ratios:  7/21*7/2

Solution:

=7/21*7/2    (7 Divided by both the numerator and denominator)

=(7*7)/(21*2)

=49/42

=7/6 or 7:6

Answer is 7/6 or 7:6

Function Examples Math

Let f be a function from X to Y.  Then it have to satisfy the following two conditions.

(i) No two different ordered pairs in f have the same first element.

(ii) All the elements of X will occur as the first elements in f.

Now let us discuss few examples that satisfies these conditions.


Example problems on functions:

Ex 1: Check whether the following ordered pair represents a function:

Here X = {1, 2, 3, 4}

Y = {a, b, c}

(i) {(1, a), (2, b), (3, c), (4, c)}.

(ii) {(1, a), (2, c), (3, b)}.

Sol: Given: (i) {(1, a), (2, b), (3, c), (4, c)}

Here all the first values are from x.  All the first values are different.  Therefore it is a function.

(ii) {(1, a), (2, c), (3, b)}.

Here 4 of X is not there in the ordered pair.  Therefore, this cannot be a function.

Ex 2: Let X = {7, 8} and Y = {1, 2}. Which one of the following is not a function of

f from X to Y.

(i) {(7, 1), (8, 2), (7, 2)}

(ii) {(7, 1), (8, 2)}

(iii) {(7, 1), (8, 1)}

Sol: (i) {(7, 1), (8, 2), (7, 2)}.  This is not a function from X to Y.  Because 7 is repeated in the first element.

(ii) {(7, 1), (8, 2)}.  This satisfies the definition of the function. Hence it is a function.

(iii) {(7, 1), (8, 1)}. This satisfies the definition of the function.  Hence it is a function.


More example problems on functions:

Ex 3: Let X = { 1,2,3,4,5,6,7,8,9} and  Y = {3,6,9}

R be a relation “is three times of”, find R on X.

Can this be a function {(3, 1), (6, 2), (9, 3)}?

Sol: R = {(3, 1), (6, 2), (9, 3)}.

Since here {(3, 1), (6, 2), (9, 3)} the first values are different, this can be a function from Y to X.

Continuous Probability Distribution

In statististics if we have the finite number of set and the probability of the set is called Discrete probability distribution. For an example for a cars on a road, deaths by cancer, tosses until a die shows a first 6. If we measure anything it is called as continuous probability distribution function. For an example we can to say the electric voltage, rainfall, and hardness of steel. In both cases we can determined the distribution by the following distribution function

F(x) = P(X ≤ x)

This is the probability that X will assume any value not exceeding x.

Continuous random variables and probability distribution:

Discrete random variables appear in experiments which is having finite set (defectives in a production, days of sun shines in Chicago, customers standing in a line etc.). Continuous random variables is appear in the experiments which we can't count but we can measure (length of screws, voltage in a power etc.). If we are find the the value probability for  the continuous  variable then it is called continuous probability distribution. By the definition of a random variable of X and its for the distribution are of continuous type will be defined as the following integral.

F(x) = `int_-oo^xf(v)dv`

it is called density of the distribution, is non negative, and is continuous perhaps except for finitely many x-values.

The continuous random variables are simpler than the discrete ones with respect to intervals. Indeed  in the continuous case the continuous probability distribution of the four probabilities corresponding to a < X ≤ b, a < X < b, a ≤ X < b, a ≤ X ≤ b with any fixed a and b (> a) are all same.

Sample problem for probability distribution:

Continuous probability distribution problem 1:

Let X have the density function f(x) = .75(1 - x2) if  -1 ≤ x ≤ 1 and zero otherwise. Find the value of distribution function. Find the continuous probability distribution of P(-1/2 ≤ X ≤ 1/2). Find x such that P(X ≤ x) = 0.95.

Solution:

F(x) = `int_-1^x`(1 - v2) dv= 0.5 + 0.75 x2 -.25x3              if  -1 ≤ x ≤ 1,

And F(x) = 1 if x > 1

P(-1/2 ≤ x ≤ 1/2) = F(1/2)  - F(-1/2) =  `int_(-1/2)^(1/2)`(0.5 + 0.75 x2 -.25x3) dx =  68.75%

Because for continuous probability distribution    P(-1/2 ≤ x ≤ 1/2) =    P(-1/2 < x ≤ 1/2)

Continuous probability distribution P(X = x) = F(x) = 0.5 + 0.75x -1.25x3  = 0.95

If we solve this we will get x = 0.73

Domain and Range of Trigonometric Functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant   Since tan ?=sin?/cos?  and sec?=1/cos?

,cos? being in denominator tan? and sec?  is not defined when cos? iszero

that is ? being odd multiple of ?/2 R- (2n + 1)/2| n`in`  Z

that is these are not defined at odd multiple of pi /2

and range of these functions is set of  all real numbers

Domain of cotangent and cosecant  since cot?= cos?/sin? and cosec?=1/sin?

sin? being in the denominator ,cot ? and cosec? is not defined when sin? is zero

that is  ? being multiple of ?

therefore for cot and cosec to be defined multiple of ? are dicarded

and range of these functions is all real numbers except interval  [-1,1]

Trignometric functions domain and range

Domain and Range of Trigonometric Functions




Explanation for domain and range of some trignometric functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant is R- {`pi` (2n + 1)/2| n `in`  Z } that is these are not defined at odd multiple of `pi`/2

and range of these functions is all values except (-1,1) interval

same can be explained  for cosecant and cotangent.

Learn Divisibility Tests

Divisibility tests means that to discover a number is divisible by a particular number (divisor) without perform the division operation. The rules shown below are used to translate a given number into a normally smaller number while preserve divisibility by the divisor. There are divisibility test for numbers in any radix, and they are all dissimilar, we represent rules simply for decimal numbers below. In this article we shall learn about divisibility tests rules with examples.

Divisibility tests rules

To learn a number is divisible by 2:

The number is end with even number (that is 0,2,4,6) the number are divisible by 2.

To learn a number is divisible by 3:

The sum of the digit is multiple of 3. The given number is divisible by 3.

To learn a number is divisible by 4:

The numbers are formed by the final pairs of digits is divisible by 4.

Example:

4672 => 72 ÷ 4 = 18

So 4672 ÷ 4 = 1168

A number is divisible by 8:

The numbers are formed by the final three digits are equally divisible by 8.

Example:

53,104 => 104 ÷ 8 = 13

So 53,104 ÷ 8 = 6638

To learn a number is divisible by 9:

The sum of the digits is a multiple of 9.

Example: 3,726

3,726 => 3 + 7 + 2 + 6 = 18

Since 18 = 9 × 2,

Then 3,726 ÷ 9 = 414

To learn a number is divisible by 6:

The numbers satisfy the rule for 2 and 3; that is, first it must be an even number, then a digit sum is a multiple of 3.

To learn a number is divisible by 12:

The numbers satisfy the rules for 3 and 4; that is, the digit sum is a multiple of 3, and its final digit pair is a multiple of 4.

Divisible tests practice problem

Problem1:

Check whether the given number 4652 is divisible by 4

Answer:

4652 => 52 ÷ 4 = 18

So 4672 ÷ 4 = 1163 therefore the given number is divisible by 4.

Problem2:

Check whether the given number 53112 is divisible by 8

Answer:

53,112 => 112 ÷ 8 = 14

So 53,112 ÷ 8 = 6639 therefore the given number is divisible by 8.

Vertical Stretch and Compression

Vertical stretch and compression mean transforming the graph based on the scale factors. Here we will see how we are performing the stretching and compression of graph using the scale factor. Stretching the graph is nothing but we are transforming the graph away from axis. Compression of the graph means squeezing the given graph towards the axis. We will see some example problems foe vertical stretch and compression of the graphs.
Vertical Stretch:

Vertical stretch is nothing but the stretching the graph away from x – axis. If the given function is f(x) then the vertical stretch of the given function is y = a f(x). Where 0 < a < 1

Example for vertical stretch:

Graph the following function and its vertical stretch. Where f(x) = x2 – 1.6x and the vertical stretch scale factor of the function f (x) is 0.4.

Solution:

Given function f(x) = x2 – 1.6x

We can write the given functions like y = x2 – 1.6x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the given function

We have to plug y = 0

So 0 = x2 – 1.6x

x (x – 1.6) = 0

x = 0 and x = 1.6

So the x intercept point is (0, 0) and (1.6, 0)

For y – intercept of the given function

We have to plug x = 0

So y = (x)2 – 1.6(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical stretch function.

The vertical stretch function is

f(x) = a f (x)

So y = 0.4 (x2 – 1.6x)

If we graph both the function we will get the following graph

vertical stretch - stretch graph
Vertical Compression:

Vertical compression is nothing but the squeezing the graph towards the x – axis. If the given function is f(x) then the vertical compression of the given function is y = af(x). Where a > 1

Example for vertical compression:

Graph the following function and its vertical compression. Where f(x) = x2 – 3.5x and the vertical compression scale factor of the function f (x) is 1.25

Solution:

The given function f(x) = x2 – 3.5x

We can write the given functions like y = x2 – 3.5x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the original function

We have to plug y = 0

So 0 = x2 – 3.5x

x (x – 3.5) = 0

x = 0 and x = 3.5

So the x intercept point is (0, 0) and (3.5, 0)

For y – intercept of the original function

We have to plug x = 0

So y = (x)2 – 3.5(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical compression function.

The vertical compression function is

f(x) = a f (x)

So y = 1.25 (x2 – 3.5x)