91Ó°ÊÓ

The imaginary unit, denoted by \(i\), is a fundamental concept in complex numbers. It is defined by the property \(i^2 = -1\). This means that \(i\) is the square root of \(-1\), something that cannot be done with real numbers alone because real numbers don't deal with negative roots.

When you see an equation involving the square root of a negative number, you can express it in terms of \(i\). For instance, \( \sqrt{-1} = i \). The imaginary unit helps us to work with otherwise impossible calculations, such as taking the square root of \(-49\) or \(-16\). By converting these into terms of \(i\), we can proceed with arithmetic operations on complex numbers.

Taking the square root of a negative number might seem tricky at first, but thanks to the imaginary unit, it becomes straightforward. When we encounter a square root of a negative number, like \( \sqrt{-49} \) or \( \sqrt{-16} \), we use the property that \(i = \sqrt{-1} \).

Here's how it works:
- \( \sqrt{-49} = \sqrt{49} \times \sqrt{-1} = 7i \)
- \( \sqrt{-16} = \sqrt{16} \times \sqrt{-1} = 4i \)
This conversion allows us to handle negative square roots conveniently within the realm of complex numbers. It's important to separate the positive root (like \( 7 \)from \( \sqrt{49} \)), and then multiply it by \( i \) to incorporate the negative sign.

Once we convert our negative square root problems into imaginary numbers, simplifying becomes much easier. Let's break it down using the original exercise:

1. Convert the negative roots:
- \( \sqrt{-49} = 7i \)
- \( \sqrt{-16} = 4i \)

2. Multiply the results:
- \( 7i \times 4i = 28i^2 \)

3. Apply the property \( i^2 = -1 \):
- Since \( i^2 = -1 \), replace \( i^2 \) in the expression:
- \( 28i^2 = 28(-1) = -28 \)

At this point, we see that the imaginary components cancel out, leaving us with a real number. The final answer must be in the form \( a + bi \), so we write:\
- \( -28 + 0i \).

Simplifying complex numbers often involves converting, multiplying, and understanding the role \( i \) plays. This process makes it clear how complex arithmetic works and allows us to solve seemingly difficult problems methodically.