![]() These are complex and complicated in nature, and thus complex number helps us to solve these problems. While analysing signals, we get quantities in terms of trigonometric functions such as sines and cosine, which vary periodically. Complex numbers are widely used in the signal analysis in the engineering field. It also tells us how electromagnetic waves travel through air and space.Ĭomplex numbers can also be used in applied fields to compute certain real-valued improper integrals employing complex-valued functions. The complex numbers help us to model or build complex electric circuits. It is employed in sub-branch of science also, in areas such as signal processing, control theory, electromagnetism, quantum mechanics, and many more. The complex numbers have real-world applications in many fields, which include physics, engineering, mathematics, and many more. But the main problem is is to get rid of that square root in the denominator.While solving problems, you must remember sometimes you have to figure out the square root \sqrt \right) ( x 1 + i y 1 ) ⋅ ( x 2 + i y 2 ) = x 1 x 2 − y 1 y 2 + i ( y 1 x 2 + y 2 x 1 ).Ī pair of complex numbers x + y i a n d x − y i x + yi \ and \ \ x - yi x + y i a n d x − y i are conjugate of each other. I find it best to simplify my numbers so I deal with smaller things. So just like we did with normal radicals, whenever we're dealing with the radical of a negative we still have to get rid of it. i squared, -1 so this just becomes -5i over 3 okay? Simplifying this out we got 5i in the numerator over 3i squared in the denominator. We have to multiply by 1, so we need an i in the top as well. So if we multiply this by i ihn the denominator, we'll get i squared, -1. Remember that i times i, i squared is -1. The 3 isn't presenting a problem, so we can leave it as this but what we really want to do is get rid of that i. Okay? So rewriting this we have 5 over 3i. So this is going to be 3i in the denominator. ![]() The first thing I want to do is to simplify that denominator radical, okay? This is square root of 9 is 3. So right here we have 5 over square root of 9. So same exact idea when we are dealing with imaginary numbers, numbers involving i. So what we ended up with is 3 root 2 over 2. When you multiply them together they just cancel each other out leaving us with what's inside which is 2. Okay? So we now have 3 root 2 in the numerator and then we have the 2 is gone away. This is going to cancel leaving me with 3. Okay.īefore I multiply that through I can see that I can simplify this. So we multiply by root 2 and then to get to the square root and square the 2 in the top as well. ![]() So now instead of having them multiply by root 8, I still need to get rid of a radical but I can multiply by root 2 instead. So what this is actually really equal to is 6 over 2 root 2. Let's do a different color so we can see it. I know that 8 is the same thing as 4 times 2. You could either multilply by root 8 over root 8 and get rid of that or what I tend to do is I like dealing with smaller numbers so if I can I try to simplify that denominator first. So whenever we're dealing with a problem like this we have to rationalize the denominator. ![]() So we're going to go back to a problem that we already know how to do. Remember that i is equal to the square root of -1 and we're not allowed to have square roots in the denominator so we have to get rid of it. So whenever we're dividing by a number that involves i, what we have to do is rationalize the denominator. Dividing by a complex number or a number involving i. ![]()
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