### Video Transcript

In this video, weβre gonna take a first look at complex and imaginary numbers. Weβll look at the definition and see some examples of complex numbers in action. Weβll solve some equations involving complex numbers and simplify some radicals and negative numbers, and this requires us to look at the definition of roots and when to use positive and negative to give two solutions.

Okay so if π₯ squared is equal to one, what is the value of π₯? Well it could be positive one, because positive one times positive one is equal to positive one. But it could also be equal to negative one, because negative one times negative one equals positive one. So there are two solutions.

Now if π₯ squared equals negative one, what would the value of π₯ be? Well Iβd need to evaluate the square root of negative one, but there isnβt a number that multiplies by itself to give negative one. Remember, positive one times positive one was equal to positive one and negative one times negative one was equal to positive one. So it looks like thereβs no solution. But since ancient Greek times in the first century A.D., mathematicians have wondered what would happen if there was a solution.

In the sixteenth century, some Italian mathematicians developed notation and rules to capture the concept of the square root of negative one, and itβs something thatβs proved very useful in many branches of mathematics ever since. It has applications as varied as electronics, signal processing, fluid dynamics, and even quantum mechanics. Itβs been really useful in designing lots of modern technology. Well we say let π squared equal negative one, so the positive or principal root of negative one is π. And we call numbers involving π either complex or imaginary numbers. Thatβs the definition. So the great imaginative leap was disabled. We donβt know what the answer is. Letβs just call it π and move on and see what happens. And as we say, it did turn out to be incredibly useful thing to do. So letβs see some examples of π in action.

Okay we want to express the square root of negative four in terms of π. Well of course, negative four is just negative one times four. And using our definition of π squared is equal to negative one, that means that π squared times four is the same as negative four. So we can split those out into the square root of π squared times the square root of four. And remember if weβve just got the square root sign, it means the principal root or the positive root, so the square root of π, squared is just π, and the square root of four is two. But we would normally express that as two π. So just using the strict definition of the radical or square root there, the square root of negative four π is equal to two π. And yes we know that two π times two π is equal to four π squared. And π squared is a negative one, so thatβs equal to negative four. And we also know that negative two π times negative two π is equal to positive four π squared which is equal to negative four as well.

So if we had started by saying that π₯ squared is equal to negative four, find all the solutions for π₯, then we would have had two solutions: positive two π or negative two π. Now letβs find an expression for the root of negative three in terms of π. Well again thatβs negative one times three makes negative three. So we can write that as π squared times three, which separates into the square root of π squared times the square root of three. Now we have to leave root three in its surd form. So that gives us an answer of root three π.

And again, thatβs one solution because of the strict meaning of root being the positive or the principal root. If weβd have started off with a question like π₯ squared equals negative three, then weβd have said that π₯ is equal to the positive or negative version of the root of negative three. And weβd have come up with two solutions: positive root three π or negative root three π. So weβve seen how π squared is defined as being negative one, and that gives us the ability to express roots of negative numbers that we couldnβt express just using the real numbers that we know and love already. And numbers that require us to use this π notation are called imaginary numbers.

Now some numbers combine real and imaginary numbers together and make complex numbers, so complex numbers have got real and imaginary components. So for example, three plus two π is a complex number. It has a real component and an imaginary component; itβs a mix of real and imaginary numbers. Another example then is five minus three π. The five is the real component and the negative three π is the imaginary component.

In general form then, complex numbers take the form π plus ππ, where π is the real part and π is the imaginary part. So this funny π symbol here is the symbol that represents complex numbers. And π and π are real numbers which tell us about the real and imaginary components of the complex number. And you can add, subtract, multiply, and divide complex numbers, and simplify the powers of π and simplify down the numbers. But we go through those in other videos, one on arithmetic and one on division of complex numbers.

Now for most people, your first encounter with complex numbers is when youβre trying to find roots of quadratic equations whose graphs donβt cut the π₯-axis. For example, π¦ equals π₯ squared minus four π₯ plus five. So if we have a look at the discriminant, π squared minus four ππ, to see whether or not thereβre gonna be any roots, we can see from the general form of our equation that in this case π is one, π is negative four, and π is five. So π squared minus four ππ is going to be negative four all squared minus four times one times five, which is sixteen minus twenty, which is obviously negative four.

So if youβve done any work with quadratics which youβve probably have if youβre watching this video, youβll know that the discriminant being less than zero means that there are no real roots. So that curve doesnβt cut the π₯-axis anywhere. There are no π₯-coordinates or π₯-values I can put into that equation to generate a π¦-coordinate of zero. But now we know a little bit about complex and imaginary numbers. Letβs go back to our quadratic formula and try to solve it. So when we substitute in our values for π, π, and π into the quadratic formula, weβve got π₯ is equal to the negative of negative four plus or minus the square root of π squared. So thatβs negative four all squared minus four times π times π four times one times five all over two π, which is two times one.

Now weβve just worked out the value of π squared minus four ππ to be negative four, so that simplifies down to four plus or minus the square root of negative four all over two. Now, before now, weβd have just shaken our heads in horror and throwing up our hands and giving up and going home at this point and said, βOh, thereβre no real roots.β But letβs apply the imaginary number rule now. Negative four can be expressed as π squared times four. And we can then split that root up into the root of π squared times the root of four. And of course the root of π squared is just π, and the root of four there is two. So weβve got four plus or minus two π over two. And we can factor the two on the top, and then the two is gonna cancel top and bottom. So weβve just got π₯ is equal to two plus or minus π.

So we have managed to find some roots, two distinct roots for this particular quadratic. Theyβre complex roots. Theyβre not real roots. So before when we were saying there are no real roots, yeah there are no real roots, but we have found two complex roots for this particular quadratic. So weβve managed to find two complex values of π₯ that when we put them into that equation, they generate a π¦-coordinate of zero. So letβs just quickly try plugging, say, π₯ equals two plus π back into the equation. So weβve got π₯ squared, thatβd be two plus π all squared minus four π₯, so thatβs minus four lots of two plus π plus five. Now letβs just check and see what that comes to and hopefully itβll come to zero.

Well the first term two plus π squared, thatβs just two plus π times two plus π. And the second term, weβre getting negative four lots of two and negative four times π. So weβre just gonna use the distributive property to work that out. So negative four times two is negative eight and negative four times π is just negative four π. And then weβve got the plus five on the end. So Iβm just gonna multiply out my brackets, so everything in the first parentheses has to be multiplied by everything in the second. So weβre gonna have two times two, two times π, π times two, and π times π. So thatβs four plus two π plus two π plus π squared. Now weβve got negative eight take away four π plus five. Now we know that π squared is negative one, so weβre adding negative one. Weβre taking away one. Thatβs the definition of π squared.

And now letβs just add up all the real parts and then add up all the imaginary parts. So weβve got four take away one take away another eight and then add five. Well that comes to zero. And weβve got two π. Add another two π, take away four π, so the result we get, the real part, is zero and the imaginary part zero. This is basically zero. So it does work, and it would also work if I did the two minus π version of π₯.

Right, letβs just quickly summarise what weβve been talking about. The definition is that π squared is equal to negative one. This means that the root of negative one is equal to π, because the root symbol there, the definition, means that weβre just taking the principle or the positive root. If we had an equation involving π₯ squared, there would be two solutions: π₯ would be positive π or π₯ would be negative π. And we defined complex numbers to have real and imaginary parts. So for example, if we said π₯ was equal to π, we could express that as a complex number: zero plus one π. Weβd have zero, real part and imaginary part of one π. Oh and just one thing before you go, watch out. Some people use π instead of π for imaginary numbers. Us mathematicians tend to use π, but engineers and electronic engineers often use π. So I hope you enjoy the whole realm of imaginary and complex numbers.