Unveiling F'(z) For F(z)=(x² - Y² + 2x) + I(2xy + 2y) Using Cauchy-Riemann Conditions

by ADMIN 86 views

Hey guys! Ever wrestled with complex functions and their derivatives? It can feel like navigating a maze, especially when those pesky Cauchy-Riemann conditions pop up. But fear not! We're about to break down a fascinating problem step-by-step, making complex differentiation feel less like a headache and more like a walk in the park. So, let's dive into this complex analysis question together and emerge victorious!

The Challenge: Finding f'(z)

We're faced with a fascinating function: f(z) = (x² - y² + 2x) + i(2xy + 2y). Our mission, should we choose to accept it (and we totally do!), is to pinpoint the correct derivative, f'(z), using the legendary Cauchy-Riemann conditions. Now, before your eyes glaze over with mathematical jargon, let's break this down. We've got a complex function, which means it has both a real part and an imaginary part. The Cauchy-Riemann conditions are our trusty tools for checking if a complex function is differentiable. If they hold true, we can then find the derivative. It's like having a secret code to unlock the derivative! This is more than just a math problem; it's a journey into the heart of complex analysis. We'll be using powerful techniques and concepts that are crucial in various fields, from physics to engineering. So, stick with me, and let's unravel this complex mystery together.

Dissecting the Function: Real and Imaginary Parts

First things first, let's clearly identify the real and imaginary parts of our function. Remember, a complex function f(z) can be expressed as f(z) = u(x, y) + iv(x, y), where u(x, y) is the real part and v(x, y) is the imaginary part. In our case, f(z) = (x² - y² + 2x) + i(2xy + 2y). So, we can see that:

  • The real part, u(x, y), is x² - y² + 2x.
  • The imaginary part, v(x, y), is 2xy + 2y.

This separation is crucial because the Cauchy-Riemann conditions hinge on the partial derivatives of these real and imaginary components. Think of it like this: we're dissecting the function to understand its inner workings. By identifying the real and imaginary parts, we're setting the stage for applying the Cauchy-Riemann conditions and ultimately finding the derivative. It's like preparing the ingredients before we start cooking a delicious mathematical dish! This step-by-step approach is key to mastering complex analysis. We're not just blindly applying formulas; we're understanding the underlying structure of the function. This deeper understanding will make the rest of the process much smoother and more intuitive.

The Cauchy-Riemann Conditions: Our Guiding Light

Now for the stars of the show: the Cauchy-Riemann conditions! These conditions are the key to unlocking the differentiability of our complex function. They're like a secret handshake that tells us if a function is "well-behaved" enough to have a derivative. The Cauchy-Riemann conditions state that for a complex function f(z) = u(x, y) + iv(x, y) to be differentiable, the following must hold true:

  1. ∂u/∂x = ∂v/∂y
  2. ∂u/∂y = -∂v/∂x

In simpler terms, the partial derivative of the real part u with respect to x must equal the partial derivative of the imaginary part v with respect to y. And, the partial derivative of u with respect to y must be the negative of the partial derivative of v with respect to x. These equations might look a bit intimidating at first, but they're actually quite elegant. They connect the real and imaginary parts of the function in a very specific way. If these conditions are satisfied, it means that the function is complex differentiable, and we can proceed to find its derivative. If they're not satisfied, then the function is not differentiable at that point. It's like a litmus test for differentiability! So, let's keep these conditions firmly in mind as we move forward. They're our guiding light in this complex analysis adventure.

Applying the Conditions: A Step-by-Step Calculation

Alright, let's get our hands dirty and apply the Cauchy-Riemann conditions to our function. This is where the rubber meets the road, guys! We'll calculate the necessary partial derivatives and see if they match up according to the conditions. Remember, our function is f(z) = (x² - y² + 2x) + i(2xy + 2y), with u(x, y) = x² - y² + 2x and v(x, y) = 2xy + 2y.

Calculating the Partial Derivatives

First, we need to find the partial derivatives of u and v with respect to x and y. This might sound complicated, but it's just a matter of applying the basic rules of differentiation. We'll treat each variable as a constant while differentiating with respect to the other. So, here we go:

  • ∂u/∂x: We differentiate u(x, y) = x² - y² + 2x with respect to x, treating y as a constant. The result is 2x + 2.
  • ∂u/∂y: We differentiate u(x, y) = x² - y² + 2x with respect to y, treating x as a constant. The result is -2y.
  • ∂v/∂x: We differentiate v(x, y) = 2xy + 2y with respect to x, treating y as a constant. The result is 2y.
  • ∂v/∂y: We differentiate v(x, y) = 2xy + 2y with respect to y, treating x as a constant. The result is 2x + 2.

See? It's not so scary when we break it down step by step. We've now got all the ingredients we need to check the Cauchy-Riemann conditions. It's like we've gathered all the pieces of a puzzle; now we just need to fit them together.

Verifying the Cauchy-Riemann Conditions

Now that we have our partial derivatives, let's plug them into the Cauchy-Riemann conditions and see if they hold true. This is the moment of truth, guys! Remember the conditions:

  1. ∂u/∂x = ∂v/∂y
  2. ∂u/∂y = -∂v/∂x

Let's substitute our calculated values:

  1. 2x + 2 = 2x + 2 (This condition holds true!)
  2. -2y = -2y (This condition also holds true!)

Hooray! Both Cauchy-Riemann conditions are satisfied. This means our function f(z) is indeed differentiable. We've successfully passed the differentiability test! It's like getting the green light to proceed to the next stage. This is a significant step because it allows us to confidently find the derivative of the function. If the conditions hadn't been satisfied, we would have known that the derivative doesn't exist. But we've cleared that hurdle, so let's move on to the grand finale: finding f'(z).

Finding the Derivative: Unveiling f'(z)

With the Cauchy-Riemann conditions happily satisfied, we're now ready to find the derivative, f'(z). This is the moment we've been working towards, guys! There are a couple of ways we can express f'(z) using our partial derivatives. The most common formulas are:

  • f'(z) = ∂u/∂x + i(∂v/∂x)
  • f'(z) = ∂v/∂y - i(∂u/∂y)

Notice that these formulas directly link the derivative to the partial derivatives we've already calculated. This is the beauty of the Cauchy-Riemann conditions; they provide a pathway to the derivative. We can choose either formula, as they will both give us the same result. Let's use the first formula for simplicity:

  • f'(z) = ∂u/∂x + i(∂v/∂x)

We already know that ∂u/∂x = 2x + 2 and ∂v/∂x = 2y. So, let's plug these values in:

  • f'(z) = (2x + 2) + i(2y)

And there we have it! We've found the derivative of our complex function. It's like reaching the summit of a mountain after a challenging climb. We've navigated the complex landscape of differentiation and emerged victorious. This result tells us how the function f(z) changes as we change the complex variable z. It's a powerful piece of information that has applications in various fields.

The Solution: Decoding the Options

Now that we've calculated f'(z) = (2x + 2) + i(2y), let's compare it to the options provided in the question. This is the final step in our journey, guys! We need to match our result with one of the given choices. The options were:

  • Option A: f'(z) = -2x + 3 - yi
  • Option B: f'(z) = 2x + xi - 4yi
  • Option C: f'(z) = 2x + 2 + 2yi

By carefully comparing our calculated derivative, f'(z) = (2x + 2) + i(2y), with the options, we can clearly see that:

  • Option C is the correct answer.

We've done it! We've successfully found the derivative of the complex function and identified the correct option. It's like solving a complex puzzle and finally fitting the last piece into place. This process demonstrates the power of the Cauchy-Riemann conditions in complex analysis. They not only tell us if a function is differentiable but also provide a direct way to calculate its derivative. This is a valuable skill to have in your mathematical toolkit.

Final Thoughts: Mastering Complex Analysis

So, there you have it! We've tackled a complex differentiation problem head-on, using the Cauchy-Riemann conditions as our guide. We've broken down the problem into manageable steps, from identifying the real and imaginary parts to calculating partial derivatives and finally finding the derivative itself. Remember, complex analysis might seem daunting at first, but with a step-by-step approach and a solid understanding of the fundamentals, you can conquer any challenge. Keep practicing, keep exploring, and you'll be a complex analysis master in no time! This journey through complex differentiation highlights the beauty and elegance of mathematics. It's not just about memorizing formulas; it's about understanding the underlying concepts and applying them creatively. So, embrace the challenge, enjoy the process, and keep learning!