Hey everyone! Welcome back to our exciting journey into the world of Complex Numbers. We've already understood what a complex number is – a beautiful blend of a real part and an imaginary part, usually written as `z = a + ib`. But what's the use of these numbers if we can't do anything with them? Just like real numbers, we need to be able to add them, subtract them, multiply them, and yes, even divide them! This is what we call the
Algebra of Complex Numbers.
Think about it: when you first learned about whole numbers, then fractions, then negative numbers, you always learned how to perform basic arithmetic operations with them, right? It's the same story with complex numbers. These operations are fundamental building blocks for everything more advanced we'll do later in complex numbers, especially for JEE!
Let's dive right in and explore how these operations work.
### 1. Addition of Complex Numbers: Adding 'Real' and 'Imaginary' Friends
Imagine you have two friends, one loves apples and the other loves bananas. When they come together, they want to know how many total apples and total bananas they have. They wouldn't try to add apples and bananas directly, would they? They'd count apples with apples and bananas with bananas.
Complex number addition works exactly the same way! You add the real parts together, and you add the imaginary parts together. It's like a component-wise addition, similar to how you add vectors!
Let's say we have two complex numbers:
`z1 = a + ib`
`z2 = c + id`
Here, `a` and `c` are the real parts, and `b` and `d` are the imaginary parts (coefficients of `i`).
To add them, `z1 + z2`:
`z1 + z2 = (a + ib) + (c + id)`
Now, group the real parts and the imaginary parts:
`z1 + z2 = (a + c) + i(b + d)`
See? Simple as that!
Key Takeaway: To add complex numbers, add their real parts and add their imaginary parts separately.
Example 1: Basic Addition
Let `z1 = 3 + 2i` and `z2 = 1 + 5i`. Find `z1 + z2`.
Solution:
Step 1: Identify the real and imaginary parts of each complex number.
For `z1`: Real part = 3, Imaginary part = 2
For `z2`: Real part = 1, Imaginary part = 5
Step 2: Add the real parts together.
`3 + 1 = 4`
Step 3: Add the imaginary parts together.
`2 + 5 = 7`
Step 4: Combine them to form the new complex number.
`z1 + z2 = 4 + 7i`
So, `(3 + 2i) + (1 + 5i) = (3+1) + i(2+5) = 4 + 7i`.
Properties of Complex Number Addition:
Just like real numbers, complex number addition has some nice properties:
- Commutative Property: The order doesn't matter. `z1 + z2 = z2 + z1`.
- Associative Property: Grouping doesn't matter. `(z1 + z2) + z3 = z1 + (z2 + z3)`.
- Additive Identity: There's a "zero" for complex numbers. It's `0 + 0i` (often just written as `0`). When you add `0 + 0i` to any complex number `z`, you get `z` back.
- Additive Inverse: For every complex number `z = a + ib`, there exists an additive inverse `-z = -a - ib` such that `z + (-z) = 0`.
### 2. Subtraction of Complex Numbers: Taking Away Parts
Subtraction is just an extension of addition, where you're adding the additive inverse. It follows the same logic as addition: subtract the real parts and subtract the imaginary parts.
Let `z1 = a + ib` and `z2 = c + id`.
To subtract `z2` from `z1`, `z1 - z2`:
`z1 - z2 = (a + ib) - (c + id)`
Distribute the minus sign carefully:
`z1 - z2 = a + ib - c - id`
Now, group the real parts and the imaginary parts:
`z1 - z2 = (a - c) + i(b - d)`
Again, straightforward!
Key Takeaway: To subtract complex numbers, subtract their real parts and subtract their imaginary parts separately.
Example 2: Basic Subtraction
Let `z1 = 5 + 7i` and `z2 = 2 - 3i`. Find `z1 - z2`.
Solution:
Step 1: Identify the real and imaginary parts.
For `z1`: Real part = 5, Imaginary part = 7
For `z2`: Real part = 2, Imaginary part = -3
Step 2: Subtract the real parts.
`5 - 2 = 3`
Step 3: Subtract the imaginary parts.
`7 - (-3) = 7 + 3 = 10`
Step 4: Combine them.
`z1 - z2 = 3 + 10i`
So, `(5 + 7i) - (2 - 3i) = (5-2) + i(7-(-3)) = 3 + i(7+3) = 3 + 10i`.
### 3. Multiplication of Complex Numbers: The 'FOIL' Method and `i² = -1`
Multiplication is where things get a little more interesting, but don't worry, it's nothing you haven't seen before. Remember multiplying two binomials in algebra, like `(x + y)(p + q)` using the FOIL method (First, Outer, Inner, Last)? We'll use the exact same approach for complex numbers, with one crucial difference: we know that
`i² = -1`. This identity is the heart of complex number multiplication!
Let `z1 = a + ib` and `z2 = c + id`.
To multiply `z1` by `z2`, `z1 * z2`:
`z1 * z2 = (a + ib)(c + id)`
Apply the FOIL method:
- First: `a * c = ac`
- Outer: `a * (id) = iad`
- Inner: `(ib) * c = ibc`
- Last: `(ib) * (id) = i²bd`
Summing these up:
`z1 * z2 = ac + iad + ibc + i²bd`
Now, substitute `i² = -1`:
`z1 * z2 = ac + iad + ibc + (-1)bd`
`z1 * z2 = ac + iad + ibc - bd`
Group the real terms and the imaginary terms:
`z1 * z2 = (ac - bd) + i(ad + bc)`
This is the general formula for multiplying two complex numbers. You don't necessarily need to memorize this formula, but you *must* remember the FOIL method and `i² = -1`. Deriving it each time is often safer and helps build understanding.
Key Takeaway: Multiply complex numbers like binomials using FOIL, and always remember to substitute `i² = -1`.
Example 3: Basic Multiplication
Let `z1 = 2 + 3i` and `z2 = 4 - 5i`. Find `z1 * z2`.
Solution:
`z1 * z2 = (2 + 3i)(4 - 5i)`
Step 1: Apply FOIL.
`First: 2 * 4 = 8`
`Outer: 2 * (-5i) = -10i`
`Inner: (3i) * 4 = 12i`
`Last: (3i) * (-5i) = -15i²`
Step 2: Combine the terms.
`8 - 10i + 12i - 15i²`
Step 3: Substitute `i² = -1`.
`8 - 10i + 12i - 15(-1)`
`8 - 10i + 12i + 15`
Step 4: Group real and imaginary parts.
`(8 + 15) + (-10 + 12)i`
`23 + 2i`
So, `(2 + 3i)(4 - 5i) = 23 + 2i`.
Properties of Complex Number Multiplication:
- Commutative Property: `z1 * z2 = z2 * z1`.
- Associative Property: `(z1 * z2) * z3 = z1 * (z2 * z3)`.
- Multiplicative Identity: The complex number `1 + 0i` (often just written as `1`) is the multiplicative identity. `z * 1 = z`.
- Distributive Property: Multiplication distributes over addition: `z1 * (z2 + z3) = z1 * z2 + z1 * z3`.
### 4. Division of Complex Numbers: The Magic of Conjugates
Division is typically the trickiest operation for beginners, but once you understand the trick, it becomes quite simple. The main goal of dividing complex numbers is to ensure that the
denominator becomes a real number. We want to remove the `i` from the bottom! How do we do that? By using something called the
complex conjugate.
First, what's a complex conjugate?
If `z = c + id` is a complex number, its conjugate, denoted as `z̄` (read as 'z bar'), is simply `c - id`. We just change the sign of the imaginary part.
Why is the conjugate so useful?
Because when you multiply a complex number by its conjugate, you always get a real number:
`(c + id)(c - id) = c² - (id)²` (This is a difference of squares: `(x+y)(x-y) = x²-y²`)
`= c² - i²d²`
`= c² - (-1)d²`
`= c² + d²`
This result, `c² + d²`, is always a real and non-negative number! Brilliant, isn't it?
So, to divide `z1` by `z2` (where `z2` is not `0 + 0i`):
`z1 / z2 = (a + ib) / (c + id)`
We multiply both the numerator and the denominator by the conjugate of the denominator (`c - id`):
`z1 / z2 = [(a + ib) * (c - id)] / [(c + id) * (c - id)]`
Let's expand the numerator (using FOIL):
`(a + ib)(c - id) = ac - iad + ibc - i²bd`
`= ac - iad + ibc + bd`
`= (ac + bd) + i(bc - ad)`
And the denominator:
`(c + id)(c - id) = c² + d²`
So, combining them:
`z1 / z2 = [(ac + bd) + i(bc - ad)] / (c² + d²)`
This can be written in the standard `A + iB` form by separating the real and imaginary parts:
`z1 / z2 = (ac + bd) / (c² + d²) + i * (bc - ad) / (c² + d²)`
Key Takeaway: To divide complex numbers, multiply the numerator and the denominator by the conjugate of the denominator. This process is similar to rationalizing the denominator when you have surds (like `1/(2 + sqrt(3))`).
Example 4: Basic Division
Let `z1 = 1 + 2i` and `z2 = 3 - i`. Find `z1 / z2`.
Solution:
Step 1: Write out the division.
`(1 + 2i) / (3 - i)`
Step 2: Find the conjugate of the denominator.
The denominator is `3 - i`. Its conjugate is `3 + i`.
Step 3: Multiply the numerator and denominator by the conjugate.
`[(1 + 2i) * (3 + i)] / [(3 - i) * (3 + i)]`
Step 4: Expand the numerator (using FOIL).
`(1 + 2i)(3 + i) = 1*3 + 1*i + 2i*3 + 2i*i`
`= 3 + i + 6i + 2i²`
`= 3 + 7i + 2(-1)`
`= 3 + 7i - 2`
`= 1 + 7i`
Step 5: Expand the denominator (complex number times its conjugate is `a² + b²`).
`(3 - i)(3 + i) = 3² + (-1)²` (Remember `i` is `0 + 1i`, so `a=3, b=-1` for `3-i`, and `a=3, b=1` for `3+i`. The form `c^2+d^2` is from `(c+id)(c-id) = c^2+d^2`, here `c=3, d=1`).
`= 9 + 1`
`= 10`
Step 6: Combine the simplified numerator and denominator.
`(1 + 7i) / 10`
Step 7: Write in `A + iB` form.
`1/10 + (7/10)i`
So, `(1 + 2i) / (3 - i) = 1/10 + (7/10)i`.
### Summary of Operations:
Here's a quick recap of the algebra rules for two complex numbers `z1 = a + ib` and `z2 = c + id`:
Operation |
Rule |
Result in `A + iB` form |
|---|
Addition |
Add real parts, add imaginary parts. |
`(a + c) + i(b + d)` |
Subtraction |
Subtract real parts, subtract imaginary parts. |
`(a - c) + i(b - d)` |
Multiplication |
FOIL method, remember `i² = -1`. |
`(ac - bd) + i(ad + bc)` |
Division |
Multiply numerator and denominator by the conjugate of the denominator. |
`[(ac + bd) / (c² + d²)] + i * [(bc - ad) / (c² + d²)]` |
###
JEE Focus Callout:
These basic operations are the bedrock of complex numbers. While direct questions on 'add these two complex numbers' are rare in JEE Mains, you will encounter these operations in almost every complex number problem. Whether it's finding the modulus of a complex expression, solving equations involving complex numbers, or working with geometrical interpretations, you'll need to be super quick and accurate with these fundamental algebraic manipulations. Make sure you practice enough so that these operations become second nature!
Keep practicing these operations, and you'll find yourself confidently navigating through more advanced topics involving complex numbers!