**Why, usually, do more complicated equations have more solutions than simple equations?**

In order to answer this question we need to learn about complex equations which are equations containing complex numbers. Before going to the question we will give a brief idea of how such types of equations are solvedand also about complex equations that involve negative square roots.

Firstly, let us discuss what is a Complex Equation?

An equation consisting of or dealing with complex numbers when one solves it is known as a complex equation. A number thatconsists of two parts, one a real part and the other part is imaginary is known as a complex number. It is written in this form:

**a+bi**

From the above, it can be seen that a and b both represent numbers whereas i stands for the imaginary part.

Let us consider an example ofa complex number.

4+3i is a complex number where 4representsthe number and the real part and 3iis said to be the imaginary part of the given number and the imaginary part is abbreviated by i.

**Imaginary Numbers**

The abbreviation i or the letter i stands for the square root of which is equal to -1.

i = √-1

Squaring the above on both sides we get -1.

The derivation of this is beyond our scope of study so one need not worry about deriving the formulas but rather just know that i =-1.

**Multiplication of Complex Numbers**

Since complex equations consist of complex numbers let us study see how the complex numbers work themselves out or let us say how the imaginary part works itself out when one needs to solve a complex equation that requires mathematical operation like the multiplication of two given complex numbers together.

Here, a brief explanation will be given.

First of all, we start with the normal multiplication step, the same way we do while multiplying two numbers or two binomials together. Considering the complex number 4+3i, firstly 4 is multiplied with each term given in the second parentheses, adding this to the other term when 3i is multiplied with each term in the given second parentheses. By doing when we multiply the two i (s) present in the equation are treated like variables as we usually do.

Let us take an example

x = (4+3i)(2+5i)

= 4(2+5i) + 3i(2+5i)

= 8 +20i + 6i +15i^{2}

= 8 + 26i +15 (-1)

= 8 + 26i – 15

= -7 +26i

With this, we can see that when we multiply the two numbers, somewhere we obtaini square which we already know has a value of -1. We can therefore replace it directly with -1, which gives us the equation as:

8 + 26i + 15 (-1) = 8 + 26i – 15 which is further simplified to:

x = -7 + 26i

Now answering the question of whether more complicated equations have more solutions than simple equations?

The answer to this can be

- The above statement can be true
- Or either false
- Or somewhere true as well as false in between.

Considering the equation below:

x−17=0.

This equation has a single solution i.e., x =17.

Let us consider another example where the equation is given as:

(x−17)(x−23)=0

This equation will have two solutions.

Again let us take into consideration another equation:

(x−17)(x−23)(x+1)(x−2–√)(x+e)=0

This equation is expected to have five solutions.

From here we can see that the more we construct polynomial equations from the roots, we will get more terms for multiplication, and hence we will tend to have more roots.Therefore from this, we can see that naturally, the morecomplicated equations get the more are the solutions. This is one that made it true.

**Do you know the answer of the below question:**

**Q: **Write equations for the following statements:

(i) The sum of numbers x and 4 is 9.

(ii) The difference between y and 2 is 8.

(iii) Ten times a is 70.

(iv) The number b divided by 5 gives 6.

(v) Three fourth of t is 15.

(vi) Seven times m plus 7 gets you 77.

(vii) One fourth of a number minus 4 gives 4.

(viii) If you take away 6 from 6 times y, you get 60.

(ix) If you add 3 to one third of z, you get 30

Now, there was once upon a time when natural numbers, or maybe the rational numbers are all we knew and that no complex numbersnor real numbers nor even negative onesexist examples like 1,2,3 and so on and their ratios yet many were able to explore different equations in those domains. Addition, subtraction, multiplication can be done easily, and find out the value of the variable x and then solve x+6 and x^{3}+17 and so on so forth. Now, polynomial equations involving integer coefficients and integer solutions can also be solved easily and the complicated ones are seen to have fewer solutions than the simpler ones.

x^{2}+y^{2}=z^{2}

The above equation has an infinite number of solutions.

Whereas, x^{23}+y^{23}=z^{23}has no solution.

There are some equations where they can be interpreted only with the help of graphs geometrically. Here, the curves in the graph will have a genus whereby the more complicated the curves the higher the genus. So, speaking in a natural context, we can say that the more complicated the equations, they may have fewer solutions.

Hence, sometimes it can be true that more complicated equations have more solutions, but sometimes the opposite might also be true.