Solving Complex Linear Equations: A Step-by-Step Guide

Hey guys, let's dive into the fascinating world of complex numbers and tackle a system of linear equations involving them! It might sound a bit intimidating at first, but trust me, with a clear approach, it's totally manageable. We're going to break down this problem piece by piece, making sure you understand every step. So, grab your favorite thinking cap, and let's get started on solving this system:

(3 - i)z₁ + (4 + 2i)z₂ = 2 + 6i (4 + 2i)z₁ - (2 + 3i)z₂ = 5 + 4i

Our goal here is to find the values of $z_1$ and $z_2$ that satisfy both of these equations simultaneously. These $z_1$ and $z_2$ are not your everyday real numbers; they are complex numbers, meaning they have a real part and an imaginary part, often written in the form $a + bi$, where $i$ is the imaginary unit ($i^2 = -1$). Dealing with complex numbers in equations follows many of the same rules as real numbers, but we need to be extra careful when performing operations like addition, subtraction, multiplication, and division with the imaginary unit $i$. Think of it like learning a new dialect of a language you already know – the structure is similar, but there are specific nuances to master.

We'll be using methods similar to those used for solving systems of real linear equations, such as substitution or elimination. The elimination method often proves quite effective when coefficients are a bit complex, as it allows us to systematically eliminate one variable. This involves multiplying one or both equations by a suitable complex number (or just a real number) to make the coefficients of either $z_1$ or $z_2$ opposites, so that when we add the equations, one variable disappears. Let's walk through this process together, making sure we handle all the complex arithmetic correctly. This journey into complex systems will not only solve our specific problem but also build your confidence in handling more advanced mathematical concepts. Remember, the key is patience and meticulous calculation. So, let's embark on this mathematical adventure, guys, and conquer these complex equations!

Understanding the Basics of Complex Numbers

Before we jump headfirst into solving our system, it's super important to have a solid grasp of how complex numbers work. This is the foundation upon which our entire solution will be built, so let's make sure we're all on the same page, guys. A complex number, generally represented as $z = a + bi$, consists of a real part ($a$) and an imaginary part ($b$). The magic happens with the imaginary unit, $i$, which is defined as the square root of -1 ($i = \sqrt{-1}$). This simple definition opens up a whole universe of numbers beyond the real number line. When we perform operations like addition and subtraction with complex numbers, we simply combine the real parts and the imaginary parts separately. For example, $(a + bi) + (c + di) = (a+c) + (b+d)i$. Similarly, $(a + bi) - (c + di) = (a-c) + (b-d)i$. Easy peasy, right?

Multiplication is where things get a little more interesting. When multiplying two complex numbers, $(a + bi)(c + di)$, we use the distributive property (like FOILing in algebra): $ac + adi + bci + bdi^2$. Now, here's the crucial part: remember that $i^2 = -1$. So, the expression becomes $ac + adi + bci - bd$. We then group the real terms and the imaginary terms: $(ac - bd) + (ad + bc)i$. This is the standard form of the product. Division is arguably the trickiest part, and it involves using the complex conjugate. The complex conjugate of a number $a + bi$ is $a - bi$. To divide $(a + bi)$ by $(c + di)$, we multiply both the numerator and the denominator by the conjugate of the denominator: $\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)}$. The denominator simplifies nicely because $(c + di)(c - di) = c^2 - (di)^2 = c^2 - d^2i^2 = c^2 - d^2(-1) = c^2 + d^2$, which is always a real number! The numerator is then expanded using the multiplication rule we just learned. So, $\frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + \frac{bc - ad}{c^2 + d^2}i$. This process ensures our final result is in the standard $a + bi$ form.

Understanding these fundamental operations is key. When we're dealing with our system of equations, these are the tools we'll be using repeatedly. We'll be multiplying complex numbers by constants, adding complex numbers, and potentially dividing by complex numbers (or their conjugates). So, take a moment to internalize these rules. If you need a quick refresher, don't hesitate to look them up. The more comfortable you are with these basic complex number arithmetic rules, the smoother our journey through solving the system will be. It's like learning the scales before playing a complex piece of music – essential for a beautiful performance!

Setting Up the Elimination Method

Alright, guys, now that we've refreshed our memory on complex number arithmetic, let's get down to business with our specific system of equations. We have:

Equation 1: $(3 - i)z_1 + (4 + 2i)z_2 = 2 + 6i$ Equation 2: $(4 + 2i)z_1 - (2 + 3i)z_2 = 5 + 4i

The elimination method is our strategy of choice here. The goal is to make the coefficients of either $z_1$ or $z_2$ in the two equations opposites. This way, when we add the equations together, that variable will be eliminated, leaving us with an equation with only the other variable. Let's choose to eliminate $z_2$. To do this, we need to make the coefficient of $z_2$ in Equation 1 the opposite of the coefficient of $z_2$ in Equation 2. Currently, we have $(4 + 2i)$ in Equation 1 and $-(2 + 3i)$ in Equation 2. These aren't easily made opposites by just multiplying by a real number.

Instead, let's try to make the coefficients of $z_1$ opposites. In Equation 1, the coefficient is $(3 - i)$, and in Equation 2, it's $(4 + 2i)$. We can try to multiply Equation 1 by some complex number $c_1$ and Equation 2 by some complex number $c_2$ such that $c_1(3 - i) = -c_2(4 + 2i)$. A simpler approach for elimination is often to multiply one equation by a factor and the other by another factor so that one variable's coefficients become equal or opposite. Let's aim to make the coefficients of $z_2$ opposites.

We have $(4 + 2i)$ in Eq 1 and $-(2 + 3i)$ in Eq 2. Let's try to manipulate Eq 2 to make its $z_2$ coefficient the negative of Eq 1's $z_2$ coefficient. To do this, we can multiply Eq 1 by $(2 + 3i)$ and Eq 2 by $(4 + 2i)$. This would make both coefficients of $z_2$ look like $(4 + 2i)(2 + 3i)$ and $-(4 + 2i)(2 + 3i)$, which are opposites. This seems like a lot of multiplication, but it's a systematic way to go.

Let's multiply Equation 1 by $(2 + 3i)$: $(2 + 3i) * [(3 - i)z₁ + (4 + 2i)z₂] = (2 + 3i) * (2 + 6i)$

This expands to: $[(2 + 3i)(3 - i)]z₁ + [(2 + 3i)(4 + 2i)]z₂ = (2 + 3i)(2 + 6i)$

Let's calculate the products: $(2 + 3i)(3 - i) = (2*3 - 3*(-1)) + (2*(-1) + 3*3)i = (6 + 3) + (-2 + 9)i = 9 + 7i$ $(2 + 3i)(4 + 2i) = (2*4 - 3*2) + (2*2 + 3*4)i = (8 - 6) + (4 + 12)i = 2 + 16i$ $(2 + 3i)(2 + 6i) = (2*2 - 3*6) + (2*6 + 3*2)i = (4 - 18) + (12 + 6)i = -14 + 18i$

So, the modified Equation 1 (let's call it Eq 1') is: $(9 + 7i)z₁ + (2 + 16i)z₂ = -14 + 18i$

Now, let's multiply Equation 2 by $(4 + 2i)$: $(4 + 2i) * [(4 + 2i)z₁ - (2 + 3i)z₂] = (4 + 2i) * (5 + 4i)$

This expands to: $[(4 + 2i)(4 + 2i)]z₁ - [(4 + 2i)(2 + 3i)]z₂ = (4 + 2i)(5 + 4i)$

Let's calculate the products: $(4 + 2i)(4 + 2i) = (4*4 - 2*2) + (4*2 + 2*4)i = (16 - 4) + (8 + 8)i = 12 + 16i$ $(4 + 2i)(2 + 3i) = (4*2 - 2*3) + (4*3 + 2*2)i = (8 - 6) + (12 + 4)i = 2 + 16i$ $(4 + 2i)(5 + 4i) = (4*5 - 2*4) + (4*4 + 2*5)i = (20 - 8) + (16 + 10)i = 12 + 26i$

So, the modified Equation 2 (let's call it Eq 2') is: $(12 + 16i)z₁ - (2 + 16i)z₂ = 12 + 26i$

Now, look at the coefficients of $z_2$ in Eq 1' and Eq 2'. We have $(2 + 16i)$ in Eq 1' and $-(2 + 16i)$ in Eq 2'. Perfect! They are opposites. This is the crucial setup for elimination. We've done the heavy lifting of complex multiplication, and the path forward is much clearer now, guys.

Solving for the First Variable ($z₁$)

We've successfully set up our equations for elimination. Here they are again:

Eq 1': $(9 + 7i)z₁ + (2 + 16i)z₂ = -14 + 18i$ Eq 2': $(12 + 16i)z₁ - (2 + 16i)z₂ = 12 + 26i$

Now comes the exciting part – elimination! We're going to add Eq 1' and Eq 2' together. Notice how the $z_2$ terms have coefficients that are negatives of each other ($+(2 + 16i)$ and $-(2 + 16i)$). When we add them, they will cancel out:

$[(9 + 7i)z₁ + (12 + 16i)z₁] + [(2 + 16i)z₂ - (2 + 16i)z₂] = (-14 + 18i) + (12 + 26i)$

Let's combine the terms:

For the $z_1$ coefficients: $(9 + 7i) + (12 + 16i) = (9 + 12) + (7 + 16)i = 21 + 23i$

The $z_2$ terms cancel out: $(2 + 16i) - (2 + 16i) = 0$

For the right-hand side: $(-14 + 18i) + (12 + 26i) = (-14 + 12) + (18 + 26)i = -2 + 44i$

So, our equation simplifies to: $(21 + 23i)z₁ = -2 + 44i$

Now, we need to isolate $z_1$. To do this, we'll divide both sides by $(21 + 23i)$:

$z₁ = \frac{-2 + 44i}{21 + 23i}$

This is a complex division problem, guys. Remember how we do this? We multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $(21 + 23i)$ is $(21 - 23i)$.

$z₁ = \frac{(-2 + 44i)(21 - 23i)}{(21 + 23i)(21 - 23i)}$

Let's calculate the numerator: $(-2 + 44i)(21 - 23i) = (-2*21 - 44*(-23)) + (-2*(-23) + 44*21)i$ $= (-42 + 1012) + (46 + 924)i$ $= 970 + 970i$

Now, let's calculate the denominator: $(21 + 23i)(21 - 23i) = 21^2 - (23i)^2 = 21^2 - 23^2i^2$ $= 441 - 529(-1)$ $= 441 + 529$ $= 970$

So, our expression for $z_1$ becomes: $z₁ = \frac{970 + 970i}{970}$

We can simplify this by dividing each term in the numerator by 970: $z₁ = \frac{970}{970} + \frac{970i}{970}$ $z₁ = 1 + 1i$

So, we found our first solution: $z₁ = 1 + i$. Pretty neat, right? We've conquered a significant part of the problem by carefully performing complex arithmetic and using the elimination method. Stick with me, guys, because the next step is to find $z_2$!

Solving for the Second Variable ($z₂$)

We've successfully found the value of $z_1$, which is $1 + i$. Now, we need to find $z_2$. The easiest way to do this is to substitute this value of $z_1$ back into one of the original equations. Let's use Equation 1, as it seems a bit simpler:

Equation 1: $(3 - i)z₁ + (4 + 2i)z₂ = 2 + 6i$

Substitute $z₁ = 1 + i$ into Equation 1:

$(3 - i)(1 + i) + (4 + 2i)z₂ = 2 + 6i$

First, let's multiply out the $(3 - i)(1 + i)$ term: $(3 - i)(1 + i) = (3*1 - (-1)*1) + (3*1 + (-1)*1)i$ $= (3 + 1) + (3 - 1)i$ $= 4 + 2i$

Now, substitute this back into our equation: $(4 + 2i) + (4 + 2i)z₂ = 2 + 6i$

Our goal is to isolate $(4 + 2i)z₂$. To do this, subtract $(4 + 2i)$ from both sides of the equation:

$(4 + 2i)z₂ = (2 + 6i) - (4 + 2i)$

Let's perform the subtraction on the right side: $(2 + 6i) - (4 + 2i) = (2 - 4) + (6 - 2)i$ $= -2 + 4i$

So now we have: $(4 + 2i)z₂ = -2 + 4i$

To find $z_2$, we need to divide both sides by $(4 + 2i)$:

$z₂ = \frac{-2 + 4i}{4 + 2i}$

This is another complex division. We'll multiply the numerator and denominator by the conjugate of the denominator, which is $(4 - 2i)$.

$z₂ = \frac{(-2 + 4i)(4 - 2i)}{(4 + 2i)(4 - 2i)}$

Let's calculate the numerator: $(-2 + 4i)(4 - 2i) = (-2*4 - 4*(-2)) + (-2*(-2) + 4*4)i$ $= (-8 + 8) + (4 + 16)i$ $= 0 + 20i$ $= 20i$

Now, let's calculate the denominator: $(4 + 2i)(4 - 2i) = 4^2 - (2i)^2 = 16 - 4i^2$ $= 16 - 4(-1)$ $= 16 + 4$ $= 20$

So, our expression for $z_2$ becomes: $z₂ = \frac{20i}{20}$

Simplify this: $z₂ = i$

And there we have it, guys! Our second solution is $z₂ = i$. We've successfully solved for both variables in our system of complex linear equations. It required careful application of complex number arithmetic and the elimination method, but the result is accurate and well-earned.

Verification of the Solution

So, we found our potential solutions: $z₁ = 1 + i$ and $z₂ = i$. But before we celebrate, it's always a good practice, especially in mathematics, to verify our solutions. This means plugging these values back into the original equations to make sure they hold true. This step ensures that we haven't made any calculation errors along the way. Think of it as a double-check to guarantee accuracy, guys.

Let's check Equation 1: $(3 - i)z₁ + (4 + 2i)z₂ = 2 + 6i$ Substitute $z₁ = 1 + i$ and $z₂ = i$: $(3 - i)(1 + i) + (4 + 2i)(i)$ We already calculated $(3 - i)(1 + i)$ in the previous step, and it resulted in $4 + 2i$. Let's calculate the second term: $(4 + 2i)(i) = 4i + 2i^2 = 4i + 2(-1) = -2 + 4i$

Now, add these two results: $(4 + 2i) + (-2 + 4i) = (4 - 2) + (2 + 4)i = 2 + 6i$

This matches the right-hand side of Equation 1 exactly! Equation 1 is satisfied. This is a great sign!

Now, let's check Equation 2: $(4 + 2i)z₁ - (2 + 3i)z₂ = 5 + 4i$ Substitute $z₁ = 1 + i$ and $z₂ = i$: $(4 + 2i)(1 + i) - (2 + 3i)(i)$ Let's calculate the first term: $(4 + 2i)(1 + i) = (4*1 - 2*1) + (4*1 + 2*1)i = (4 - 2) + (4 + 2)i = 2 + 6i$

Now, let's calculate the second term: $(2 + 3i)(i) = 2i + 3i^2 = 2i + 3(-1) = -3 + 2i$

Now, perform the subtraction (remember the minus sign in front of the second term): $(2 + 6i) - (-3 + 2i) = (2 - (-3)) + (6 - 2)i$ $= (2 + 3) + (4)i$ $= 5 + 4i$

This also matches the right-hand side of Equation 2 exactly! Equation 2 is satisfied.

Since both original equations are satisfied with our values of $z_1$ and $z_2$, we can be absolutely confident that our solution is correct. The solutions are $z₁ = 1 + i$ and $z₂ = i$. We've successfully navigated the complexities of these equations, guys, and emerged with the correct answers. This verification step is crucial for building trust in your mathematical work and ensuring accuracy. Great job, everyone!

Conclusion

And there you have it, guys! We've successfully tackled a system of linear equations involving complex unknowns. We started by understanding the fundamental rules of complex number arithmetic – addition, subtraction, multiplication, and division, paying special attention to the role of the imaginary unit $i$ and complex conjugates. Then, we strategically applied the elimination method, which involved multiplying our original equations by carefully chosen complex numbers to make the coefficients of one variable opposites. This allowed us to eliminate $z_2$ and solve for $z_1$. Once we had $z_1$, we substituted it back into one of the original equations to solve for $z_2$. Finally, and most importantly, we rigorously verified our solutions by plugging them back into both original equations, confirming that they indeed satisfied the system. The solutions we found are $z₁ = 1 + i$ and $z₂ = i$.

Solving systems with complex numbers might seem like a step up in difficulty, but as we've demonstrated, it's a logical extension of solving real systems. The core principles remain the same: aim to isolate variables and use algebraic manipulation. The added layer is simply the careful handling of complex number operations. Remember, practice is key! The more you work with complex numbers, the more intuitive their arithmetic will become. Don't be afraid to go back and review the steps if you ever feel unsure. This kind of problem-solving builds critical thinking and a deeper appreciation for the structure of mathematics. Keep practicing, keep exploring, and you'll master these concepts in no time. Well done, everyone, on sticking through this detailed walkthrough!