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**Complex Numbers**

Let us try and solve the equation x^{2}+1 =0. We can simplify it and write it as x^{2}=-1 or x = ± √−1. But what is the root of -1? It is not on the number line. So does it simply not exist? Well in mathematics, sticking to reality has never been a priority! This solution does not lie on the number line, that means it must be off it then.

We can write -a = (-1)×(a), where ‘a’ is a real number. The square root of a negative number can be written as

Let us use a symbol for √−1. Let’s denote it by ‘i’ ( from iota, Greek for extremely small). Then i = √−1 is a number that doesn’t fall anywhere on the number line! Such numbers that are not on the real number line, are the imaginary numbers. They are also known as the complex numbers.

“Now wait for a second, we only have ‘i’, why are you calling it numbers? Where are the others?” We can get every real number from other real numbers by certain algebraic operations. Like if you keep on adding the number one to itself, you will get the set of natural numbers and so on.

Similarly, all the numbers that have ‘i’ in them are the imaginary numbers. The number i, is the imaginary unit. 3i, 4i, -i, √−9 etc. are all imaginary numbers. Let us have a general definition of the imaginary or complex numbers.

**Why So Complex!**

Let ‘a’ , ‘b’ be two real numbers. Let i = √−1, then any number of the form a + ib is a complex number. The reason to define a complex number in this way is to make a connection between the real numbers and the complex ones. For example, we can write, 2 = 2 + 0.i. Therefore, every real number can be written in the form of a + ib; where b = 0.

Also if a complex number is such that a = 0, we call it a purely imaginary number. In general, a is known as the “real” part and b is known as the “imaginary” or the complex part of the imaginary number. “Why am I doing this again?” Well, hold on. Let’s recall the equation that we couldn’t quite figure out the solution for i.e. x^{2}+1=0

Hence, x = ± √−1 or x = ± i. Substituting for x, in the equation we have (i)^{2}+1=0

i^{2} =±[√−1]^{2}=-1, using the above equation we have, (-1)+1=0 satisfies the equation. Substitution of -i also gives the same results. So there you go, now you can solve equations that you would have rather just left alone. To sum up, all the numbers of the form a +ib, where a and b are real and i =√−1 , are called imaginary numbers. We usually denote an imaginary number by ‘z’.

**Equality of Complex Numbers**

Let us practice the concepts we have read this far.**Example 1:** There are two numbers z_{1} = x + iy and z_{2} = 3 – i7. Find the value of x and y for z_{1} = z_{2}. What is the sum of Re (z_{1}, z_{2})?**Solution:** We have z_{1} = x + iy and z_{2} = 3 – i7

First of all, real part of any complex number (a+ib) is represented as Re(a + ib) = a and imaginary part of (a +ib) is represented as Im(a+ib) = b. Also, when two complex numbers are equal, their corresponding real parts and imaginary parts must be equal. Therefore, if a + ib = c + id, then Re(a+ib) = Re(c+id) and Im(a+ib) = Im(c+id)

Conversely if two complex numbers say z1 and z2, are such that Re(z_{1}) = Re(z_{2}) and Im(z_{1}) = Im(z_{2}), then z_{1}=z_{2}. In other words, we must have a = c and b = d. We can’t have a = d because we can’t relate the real and the imaginary parts together. Let us come back to our problem.

z_{1 }= z_{2} or x + iy = 3 – i7

Re(x + iy) = x Re(3 – i7) = 3

Im(x + iy) = y Im(3 – i7) = -7

This means x = 3 and y = -7. Therefore, Re(z_{1},z_{2}) = 3,3 and hence, the sum of real parts of z1 and z2 = 3 + 3 = 6.

Integer Powers of ‘i’

As we saw already, i2 = -1. Also, i3 = i2(i) = -i. Similarly, you can find any given power of i, by reducing it to the above two forms.

**More Solved Examples For You****Example 2:** If 2i^{2}+6i^{3}+3i^{16}−6i^{19}+4i^{25}=x+iy, then (x,y)= ?**A)** (1 , 4)**B)** (4 , 1)**C)** (-1 , 4)**D)** (-1 , -4)**Solution: (A)** The above expression can be simplified as:

2 (-1) + 6 (-i) + 3 ( i )^{2×2×2×2 }– 6 (i^{18} . i) + 4 (i^{24}.i) = x + iy

Therefore, -2 – 6i + 3 (1) – 6 (-1)^{9}.i + 4i = x + iy Or 1 + 4i = x + iy

Therefore, x = 1 and y = 4 or (x,y) = (1,4)

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