Table of contents | |
Complex Numbers - Introduction | |
Why So Complex! | |
Equality of Complex Numbers | |
More Solved Examples For You |
Let us try and solve the equation x2+1 =0. We can simplify it and write it as x2=-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!
“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.
Let us have a general definition of the imaginary or complex numbers.
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. x2+1=0.
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’.
Let us practice the concepts we have read this far.
Example 1: There are two numbers z1 = x + iy and z2 = 3 – i7. Find the value of x and y for z1 = z2. What is the sum of Re (z1, z2)?
Solution: We have z1 = x + iy and z2 = 3 – i7.
Example 2: If 2i2+6i3+3i16−6i19+4i25=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 (i18 . i) + 4 (i24.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)
Example 3: Perform the indicated operation and write your answer in standard form.
(4−5i)(12+11i)
Solution: We know how to multiply two polynomials and so we also know how to multiply two complex numbers. All we need to do is “foil” the two complex numbers to get,
(4−5i)(12+11i) = 48+44i−60i−55i2 = 48−16i−55i2
All we need to do to finish the problem is to recall that i2=−1. Upon using this fact we can finish the problem.
(4−5i)(12+11i)=48−16i−55(−1)=103−16i
161 videos|58 docs
|
1. What are complex numbers? |
2. How are complex numbers represented graphically? |
3. What does it mean for two complex numbers to be equal? |
4. How can complex numbers be added or subtracted? |
5. How can complex numbers be multiplied or divided? |
|
Explore Courses for Mathematics exam
|