Master Complex Number Powers: Easy Calculations

Hey math whizzes! Today, we're diving deep into the fascinating world of complex numbers and how to tackle those intimidating power calculations. You know, the ones that look like (1+i)²⁵ or something even crazier? Don't sweat it, guys! We're going to break down these problems step-by-step, making them super manageable and, dare I say, even fun.

We'll be using some awesome tools like De Moivre's Theorem, which is like a secret weapon for dealing with powers of complex numbers. It might sound fancy, but trust me, it's a game-changer. By the end of this, you'll be calculating these powers like a pro, impressing your friends and teachers alike. So grab your calculators, a comfy seat, and let's get this math party started!

Understanding Complex Numbers and Powers

Before we jump into the nitty-gritty calculations, let's quickly recap what complex numbers are and why dealing with their powers can be tricky. A complex number is basically a number that can be expressed in the form a + bi, where 'a' is the real part and 'b' is the imaginary part, and 'i' is the imaginary unit, where i² = -1. Think of them as existing on a 2D plane, not just a number line. This 'i' is what makes things interesting – and sometimes a little confusing!

Now, when we talk about powers of complex numbers, we're essentially multiplying a complex number by itself a certain number of times. For example, (1+i)² means (1+i) * (1+i). Doing this manually for high powers, like (1+i)²⁵, would be an absolute nightmare. Imagine multiplying (1+i) by itself 25 times – no thanks! That's where mathematical theorems come in handy, saving us tons of time and preventing potential mental meltdowns. De Moivre's Theorem is our superhero in this scenario. It provides a straightforward way to compute powers of complex numbers when they are expressed in polar form (r(cos θ + i sin θ)). The theorem states that

[r(cos θ + i sin θ)]ⁿ = rⁿ(cos(nθ) + i sin(nθ))

This is HUGE, guys! It transforms a complex multiplication problem into a much simpler calculation involving the modulus (r) and the argument (θ). We'll be converting our complex numbers into this polar form, applying the theorem, and then converting back if needed. It's a systematic approach that makes even the most daunting exponents manageable. So, the first step is always to recognize if a direct calculation is feasible or if we need to bring out the big guns like De Moivre's Theorem. For simple powers like 2 or 3, direct multiplication might be okay, but once you hit exponents like 15, 20, or 25, it's De Moivre's time to shine. We'll ensure that we cover how to find the modulus (distance from origin) and argument (angle with the positive x-axis) for our complex numbers, as these are crucial for using the theorem effectively. Remember, complex number calculations become much simpler with the right tools, and De Moivre's Theorem is definitely one of them. We are looking for ways to simplify these expressions efficiently, and understanding the underlying principles is key. So, let's get ready to apply these concepts to our specific problems.

Problem a) Calculating (1+i)²⁵

Alright, let's tackle the first beast: calculate (1+i)²⁵. This is where De Moivre's Theorem really shows its power. First off, we need to convert our complex number, (1+i), into its polar form. The general polar form of a complex number is r(cos θ + i sin θ), where r is the modulus and θ is the argument.

For (1+i):

• Modulus (r): This is the distance from the origin (0,0) to the point (1,1) on the complex plane. We calculate it using the Pythagorean theorem: r = √(a² + b²). So, for (1+i), r = √(1² + 1²) = √2.
• Argument (θ): This is the angle the line segment from the origin to (1,1) makes with the positive real axis. Since both the real and imaginary parts are positive (1 and 1), it's in the first quadrant. We can find the angle using the arctangent function: θ = arctan(b/a). So, θ = arctan(1/1) = arctan(1) = π/4 radians (or 45 degrees).

So, the polar form of (1+i) is √2(cos(π/4) + i sin(π/4)).

Now, we apply De Moivre's Theorem:

[r(cos θ + i sin θ)]ⁿ = rⁿ(cos(nθ) + i sin(nθ))

In our case, r = √2, θ = π/4, and n = 25.

(1+i)²⁵ = [√2(cos(π/4) + i sin(π/4))]²⁵

= (√2)²⁵ * (cos(25 * π/4) + i sin(25 * π/4))

Let's break down the parts:

• (√2)²⁵: This is the same as (2^(1/2))²⁵ = 2^(25/2). We can rewrite this as 2¹² * 2^(1/2) = 2¹²√2.
• cos(25π/4) + i sin(25π/4): To simplify the angle 25π/4, we can find an equivalent angle within 0 to 2π. 25π/4 = (24π + π)/4 = 6π + π/4. Since 6π represents three full rotations (6π = 3 * 2π), the angle 25π/4 is equivalent to π/4. So, cos(25π/4) = cos(π/4) = √2/2 And sin(25π/4) = sin(π/4) = √2/2

Now, let's put it all back together:

(1+i)²⁵ = 2¹²√2 * (√2/2 + i √2/2)

= 2¹²√2 * (√2/2) + 2¹²√2 * (i √2/2)

= 2¹² * (√2 * √2) / 2 + i * 2¹² * (√2 * √2) / 2

= 2¹² * 2 / 2 + i * 2¹² * 2 / 2

= 2¹² * 1 + i * 2¹² * 1

= 2¹² + i 2¹²

We can factor out 2¹²:

= 2¹²(1 + i)

And there you have it! The answer to (1+i)²⁵ is 2¹²(1+i). See? Not so scary after all when you use the right techniques. This is a classic example of how De Moivre's Theorem simplifies complex number exponentiation significantly.

Problem b) Calculating ((1+i√3)/(1−i))²⁰

Next up, we've got a slightly more complex expression: calculate ((1+i√3)/(1−i))²⁰. This one involves division first, then exponentiation. The key here is to simplify the fraction inside the parentheses before raising it to the power of 20. We can do this by converting both the numerator and the denominator to their polar forms, dividing them, and then applying De Moivre's Theorem.

Let's break it down:

Step 1: Convert the numerator (1 + i√3) to polar form.

• Modulus (r₁): r₁ = √(1² + (√3)²) = √(1 + 3) = √4 = 2.
• Argument (θ₁): θ₁ = arctan(√3/1) = arctan(√3) = π/3 (since it's in the first quadrant). So, 1 + i√3 = 2(cos(π/3) + i sin(π/3)).

Step 2: Convert the denominator (1 - i) to polar form.

• Modulus (r₂): r₂ = √(1² + (-1)²) = √(1 + 1) = √2.
• Argument (θ₂): θ₂ = arctan(-1/1) = arctan(-1). Since the real part is positive (1) and the imaginary part is negative (-1), it's in the fourth quadrant. The principal value for arctan(-1) is -π/4. So, θ₂ = -π/4 (or 7π/4). So, 1 - i = √2(cos(-π/4) + i sin(-π/4)).

Step 3: Divide the complex numbers in polar form.

To divide complex numbers in polar form, we divide their moduli and subtract their arguments:

(r₁ (cos θ₁ + i sin θ₁)) / (r₂ (cos θ₂ + i sin θ₂)) = (r₁/r₂) * (cos(θ₁ - θ₂) + i sin(θ₁ - θ₂))

So, (1 + i√3) / (1 - i) = (2/√2) * (cos(π/3 - (-π/4)) + i sin(π/3 - (-π/4)))

= √2 * (cos(π/3 + π/4) + i sin(π/3 + π/4))

Let's find a common denominator for the angles: π/3 + π/4 = 4π/12 + 3π/12 = 7π/12.

So, (1 + i√3) / (1 - i) = √2 * (cos(7π/12) + i sin(7π/12)).

Step 4: Apply De Moivre's Theorem to the result.

Now we need to raise this result to the power of 20:

[√2 * (cos(7π/12) + i sin(7π/12))]²⁰

= (√2)²⁰ * (cos(20 * 7π/12) + i sin(20 * 7π/12))

• (√2)²⁰: This is (2^(1/2))²⁰ = 2¹⁰.
• cos(20 * 7π/12) + i sin(20 * 7π/12): Let's simplify the angle: 20 * 7π/12 = 140π/12 = 35π/3. To find an equivalent angle, we can subtract multiples of 2π (or 6π/3). 35π/3 = (30π + 5π)/3 = 10π + 5π/3. 10π represents 5 full rotations, so the angle is equivalent to 5π/3. Alternatively, 35π/3 = (36π - π)/3 = 12π - π/3. This means the angle is equivalent to -π/3. Let's use -π/3 as it's simpler. So, cos(35π/3) = cos(-π/3) = cos(π/3) = 1/2. And sin(35π/3) = sin(-π/3) = -sin(π/3) = -√3/2.

Step 5: Put it all together.

((1+i√3)/(1−i))²⁰ = 2¹⁰ * (1/2 + i (-√3/2))

= 2¹⁰ * (1/2) - i * 2¹⁰ * (√3/2)

= 2⁹ * 1 - i * 2⁹ * √3

= 2⁹ - i 2⁹√3

Factor out 2⁹:

= 2⁹(1 − i√3)

And that's the answer for part (b)! We successfully navigated the division and exponentiation using polar forms and De Moivre's Theorem. This demonstrates how breaking down complex problems into smaller, manageable steps is crucial for complex number manipulation.

Problem c) Combining Powers: A Challenging Finale

Finally, we face the most intricate problem: calculate ((−1+i√3)¹⁵/(1−i)²⁰) + ((−1−i√3)¹⁵/(1+i)²⁰. This problem combines elements from the previous two, requiring us to calculate two separate complex fractions raised to powers and then add them together. It's a true test of our understanding of complex number arithmetic and De Moivre's Theorem.

Let's tackle each fraction separately.

Part 1: Calculate ((−1+i√3)¹⁵/(1−i)²⁰)

We need to find the polar forms of the complex numbers involved.

Numerator: (−1 + i√3)

• Modulus (r₃): r₃ = √((-1)² + (√3)²) = √(1 + 3) = √4 = 2.
• Argument (θ₃): The point (-1, √3) is in the second quadrant. The reference angle is arctan(√3/|-1|) = arctan(√3) = π/3. Since it's in the second quadrant, θ₃ = π - π/3 = 2π/3. So, −1 + i√3 = 2(cos(2π/3) + i sin(2π/3)).

Now, raise it to the power of 15: [2(cos(2π/3) + i sin(2π/3))]¹⁵ = 2¹⁵ * (cos(15 * 2π/3) + i sin(15 * 2π/3)) = 2¹⁵ * (cos(30π/3) + i sin(30π/3)) = 2¹⁵ * (cos(10π) + i sin(10π)) Since 10π is a multiple of 2π, cos(10π) = 1 and sin(10π) = 0. So, (−1 + i√3)¹⁵ = 2¹⁵ * (1 + i * 0) = 2¹⁵.

Denominator: (1 - i)²⁰

From Problem (b), we know the polar form of (1 - i) is √2(cos(-π/4) + i sin(-π/4)). Raise it to the power of 20: [√2(cos(-π/4) + i sin(-π/4))]²⁰ = (√2)²⁰ * (cos(20 * -π/4) + i sin(20 * -π/4)) = 2¹⁰ * (cos(-5π) + i sin(-5π)) Since -5π is coterminal with -π (or π), cos(-5π) = -1 and sin(-5π) = 0. So, (1 - i)²⁰ = 2¹⁰ * (-1 + i * 0) = -2¹⁰.

Now, divide the numerator by the denominator: (−1+i√3)¹⁵ / (1−i)²⁰ = 2¹⁵ / (-2¹⁰) = -2^(15-10) = -2⁵ = -32.

Part 2: Calculate ((−1−i√3)¹⁵/(1+i)²⁰)

Let's find the polar forms for this part.

Numerator: (−1 − i√3)

• Modulus (r₄): r₄ = √((-1)² + (-√3)²) = √(1 + 3) = √4 = 2.
• Argument (θ₄): The point (-1, -√3) is in the third quadrant. The reference angle is arctan(|-√3|/|-1|) = arctan(√3) = π/3. Since it's in the third quadrant, θ₄ = π + π/3 = 4π/3. (Alternatively, we can use -2π/3). So, −1 − i√3 = 2(cos(4π/3) + i sin(4π/3)).

Now, raise it to the power of 15: [2(cos(4π/3) + i sin(4π/3))]¹⁵ = 2¹⁵ * (cos(15 * 4π/3) + i sin(15 * 4π/3)) = 2¹⁵ * (cos(60π/3) + i sin(60π/3)) = 2¹⁵ * (cos(20π) + i sin(20π)) Since 20π is a multiple of 2π, cos(20π) = 1 and sin(20π) = 0. So, (−1 − i√3)¹⁵ = 2¹⁵ * (1 + i * 0) = 2¹⁵.

Denominator: (1 + i)²⁰

From Problem (a), we know the polar form of (1 + i) is √2(cos(π/4) + i sin(π/4)). Raise it to the power of 20: [√2(cos(π/4) + i sin(π/4))]²⁰ = (√2)²⁰ * (cos(20 * π/4) + i sin(20 * π/4)) = 2¹⁰ * (cos(5π) + i sin(5π)) Since 5π is coterminal with π, cos(5π) = -1 and sin(5π) = 0. So, (1 + i)²⁰ = 2¹⁰ * (-1 + i * 0) = -2¹⁰.

Now, divide the numerator by the denominator: (−1−i√3)¹⁵ / (1+i)²⁰ = 2¹⁵ / (-2¹⁰) = -2^(15-10) = -2⁵ = -32.

Part 3: Add the results

Now we just add the results from Part 1 and Part 2:

(-32) + (-32) = -64

And that's our final answer for part (c)! This problem really tested our ability to handle multiple complex number operations including powers, division, and addition. It shows how important it is to be meticulous with each step, especially when dealing with angles and their periodicity.

Conclusion: You've Got This!

So there you have it, guys! We've conquered some seriously intimidating complex number power problems. We've seen how De Moivre's Theorem is an absolute lifesaver, transforming messy multiplications and divisions into straightforward calculations. Remember the key steps: convert to polar form (find modulus and argument), apply De Moivre's Theorem (raise modulus to the power, multiply argument by the power), and then simplify. For fractions, divide the moduli and subtract the arguments before applying the power.

These calculations might seem daunting at first, but with practice, they become second nature. Don't be afraid to break down complex expressions into smaller parts, just like we did with problem (c). Every math problem is just a series of smaller, manageable steps waiting to be solved. Keep practicing, keep exploring, and you'll find that complex numbers are not only calculable but also incredibly elegant!

Keep up the great work, and happy calculating!