Simplify Complex Number Division: (16+61i) / (4+5i)

Hey math whizzes and number nerds! Ever feel like diving into the world of complex numbers is like trying to solve a puzzle? Well, you're in the right place, guys! Today, we're going to tackle a specific type of problem: complex number division. Specifically, we're going to simplify this beast: $(16+61i) ext{ ÷ } (4+5i)$. Now, I know what you might be thinking – "Division? With imaginary numbers? That sounds tricky!" But trust me, once you get the hang of it, it's actually pretty straightforward. We'll break it down step-by-step, so by the end of this, you'll be dividing complex numbers like a pro. So grab your favorite thinking cap, maybe a cup of coffee, and let's get this done!

Understanding Complex Numbers and Division

Alright, let's kick things off by getting on the same page about complex numbers and what it means to divide them. Remember, a complex number has two parts: a real part and an imaginary part. It's usually written in the form $a + bi$, where '$a$' is the real part and '$b$' is the imaginary part, and '$i$' is the imaginary unit, which is equal to the square root of -1. So, when we talk about dividing complex numbers, like in our problem $(16+61i) ext{ ÷ } (4+5i)$, we're essentially trying to find a new complex number, let's call it $c + di$, such that when you multiply it by the divisor $(4+5i)$, you get the dividend $(16+61i)$. It's like asking, "What number, when multiplied by $(4+5i)$, gives us $(16+61i)$?"

The key to dividing complex numbers lies in a clever trick involving something called the complex conjugate. For any complex number in the form $a + bi$, its complex conjugate is $a - bi$. It's like a mirror image, where you just flip the sign of the imaginary part. Why is this conjugate so important, you ask? Well, when you multiply a complex number by its conjugate, something magical happens: the imaginary part disappears, and you're left with a real number! Specifically, $(a+bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 - b^2(-1) = a^2 + b^2$. See? No more '$i$'! This property is our secret weapon for simplifying complex number division. We use it to get rid of the complex number in the denominator, turning our division problem into a multiplication problem that we already know how to solve.

So, the general strategy for dividing $(a+bi) ext{ ÷ } (c+di)$ is to multiply both the numerator $(a+bi)$ and the denominator $(c+di)$ by the complex conjugate of the denominator, which is $(c-di)$. This keeps the value of the fraction the same (because we're multiplying by 1 in the form of rac{c-di}{c-di}), but it transforms the denominator into a nice, neat real number. Then, you just expand and simplify the numerator and the denominator separately, and finally, express the result in the standard $x+yi$ form. It might sound like a few extra steps, but it's the most reliable way to get to the correct, simplified answer. Let's get to our specific example now!

Step-by-Step Division of $(16+61i)$ by $(4+5i)$

Alright, let's dive headfirst into our actual problem: dividing complex numbers $(16+61i) ext{ ÷ } (4+5i)$. Remember our strategy? We need to multiply both the numerator and the denominator by the complex conjugate of the denominator. Our denominator is $(4+5i)$, so its complex conjugate is $(4-5i)$. Got it? Perfect!

So, our problem now looks like this:

$$\frac{16+61i}{4+5i} \times \frac{4-5i}{4-5i}$$

This might seem like we're making it more complicated, but trust the process, guys! We're doing this to eliminate that pesky '$i$' in the denominator. Now, we need to perform two multiplications: one in the numerator and one in the denominator.

Numerator Multiplication: We multiply $(16+61i)$ by $(4-5i)$. We'll use the FOIL method (First, Outer, Inner, Last) here:

• First: $16 imes 4 = 64$
• Outer: $16 imes (-5i) = -80i$
• Inner: $61i imes 4 = 244i$
• Last: $61i imes (-5i) = -305i^2$

Now, let's combine these terms. Remember that $i^2 = -1$. So, $-305i^2$ becomes $-305(-1) = 305$.

Putting it all together: $64 - 80i + 244i + 305$

Combine the real parts ($64 + 305$) and the imaginary parts ($-80i + 244i$):

• Real part: $64 + 305 = 369$
• Imaginary part: $-80i + 244i = 164i$

So, the simplified numerator is $369 + 164i$. Pretty neat, right?

Denominator Multiplication: Now, we multiply the original denominator $(4+5i)$ by its conjugate $(4-5i)$. We can use the shortcut we talked about earlier: $(a+bi)(a-bi) = a^2 + b^2$. In this case, $a=4$ and $b=5$.

So, $(4+5i)(4-5i) = 4^2 + 5^2$

• $4^2 = 16$
• $5^2 = 25$

Adding them together: $16 + 25 = 41$.

Look at that! The denominator is just a plain old real number, 41. No imaginary part to worry about!

Simplifying the Result to Standard Form

We've done the heavy lifting, guys! We've multiplied our original fraction by the conjugate pair, and now we have:

$$\frac{369 + 164i}{41}$$

The final step in complex number division is to express this result in the standard form $x+yi$. To do this, we simply divide both the real part and the imaginary part of the numerator by the denominator.

So, we split our fraction into two parts:

$$\frac{369}{41} + \frac{164i}{41}$$

Now, we just need to simplify these two fractions. Let's check if 369 is divisible by 41. A quick calculation shows that $369 ext{ ÷ } 41 = 9$. Perfect!

Next, let's check if 164 is divisible by 41. We can try multiplying 41 by small integers. $41 imes 1 = 41$, $41 imes 2 = 82$, $41 imes 3 = 123$, $41 imes 4 = 164$. Bingo! So, $164 ext{ ÷ } 41 = 4$.

Now, we can write our final answer in the standard $x+yi$ form:

$$9 + 4i$$

And there you have it! We've successfully divided and simplified the complex numbers $(16+61i) ext{ ÷ } (4+5i)$ to get $9+4i$. Wasn't so bad, right? It just takes a little practice with the complex conjugate trick. Remember, the key is always to get rid of that '$i$' in the denominator by multiplying by its conjugate. Keep practicing, and you'll master complex number division in no time! Keep exploring the fascinating world of mathematics, and I'll catch you in the next one!