 This is a video about further complex numbers. I'm going to be looking at some questions from the review exercise in our FP2 textbook. The first question I'm going to look at is number 3. It tells us that the equation z cubed plus 27i equals 0 has the complex roots alpha, beta and gamma. We're told to find the roots, giving our answers an exact Cartesian form. Now by far the best way of answering a question like this is to write z in modulus argument form. So we'll say that z is equal to r e to the i theta. It follows from that the cube of r e to the i theta plus 27i is equal to 0. And now we can solve this equation to find out what r and theta are. First of all we'll say that r cubed times e to the power of 3i theta plus 27i equals 0. And then that r cubed e to the 3i theta is equal to minus 27i. Now minus 27i is simply the complex number with modulus 27 and argument 3pi over 2. So we can write that as 27 times e to the power of i times 3pi over 2. Now if you compare the two sides of this equation it becomes obvious that r cubed must be 27 and 3 theta must be 3pi over 2 or 3pi over 2 plus a multiple of 2pi. Because remember if you add a multiple of 2pi to the argument of a complex number it makes no difference. You've still got the same complex number. So 3 theta could be 3pi over 2 or 7pi over 2 or 11pi over 2 and so on. Okay so it must be that r is 3 and theta could be pi over 2 or 7pi over 6 or 11pi over 6 or whatever. So now we know what r and theta are. So we know the modulus and the argument of z. All that remains is to write z in exact Cartesian form. So remember the modulus must be 3 and the argument could be pi over 2 or 7pi over 6 or 11pi over 6 or whatever. Now the first possibility is that theta is pi over 2 in which case you've got the number 3 times e to the i pi by 2. But that's just the same as 3i. The complex number with modulus 3 and argument pi over 2 is obviously 3i. The next possibility is that theta is 7pi over 6. So in that case we're dealing with 3e to the i 7pi by 6. And to find that in Cartesian form we'll need to say that that's the same as 3 times the cosine of 7pi over 6 plus i times the sine of 7pi over 6. Remember that that's the same as 3 times minus the cosine of pi by 6 minus i times the sine of pi by 6. And so therefore it's equal to 3 lots of minus root 3 over 2 minus i over 2 or multiply it out minus 3 root 3 over 2 minus 3i over 2. So that's the second possible value of z. The third possible value of z comes from taking theta equals 11pi over 6 and then we're dealing with the number 3e to the i 11pi over 6 which can be written 3 times the cosine of 11pi over 6 plus i times the sine of 11pi over 6 or as 3 times the cosine of pi over 6 minus i times the sine of pi over 6. Which turns out to be 3 root 3 over 2 minus 3i over 2. So here are the three possible values of z in exact Cartesian form. And so alpha, beta and gamma are these three numbers in some order. The next part of this question tells us to show that alpha plus beta plus gamma equals zero and to use that result to find the complex number alpha plus beta cubed. Well we can show that alpha plus beta plus gamma equals zero simply by adding up our answers. If you add them up you'll see that it cancels to 3i minus 3i over 2 minus 3i over 2 which comes to zero. So that's easy. The next bit is a little trickier. Since alpha plus beta plus gamma equals zero we can say that alpha plus beta is equal to minus gamma. And now we can cube both sides. The cube of alpha plus beta must be the cube of minus gamma. But the cube of minus gamma will be minus the cube of gamma. The whole point of gamma is that its cube is minus 27i. So this is minus minus 27i or simply 27i. So therefore the cube of alpha plus beta is 27i. Another part of the question asks us to verify that minus a third of i alpha is a cube root of unity. And then we've got to use this fact to work out the value of one minus a third of i alpha minus a ninth of alpha squared. Now to show that something is a cube root of unity we have to show that its cube is equal to one. The task here is to show that the cube of minus a third i alpha is one. Well if we cube the bit separately we've got the cube of minus a third times the cube of i times alpha cubed. And that's going to be minus one over 27 times minus i times minus 27i. And if you rearrange the pieces of that you can see that that's the same as one times one which is simply one. That proves that minus a third of i alpha is a cube root of unity. For the next part you need to spot that minus a ninth of alpha squared is the square of minus a third of i alpha. Let's check that. The square of minus a third i alpha be the square of minus a third times the square of i times the square of alpha which is going to be a ninth times minus one times alpha squared. So it is indeed minus a ninth of alpha squared. Now that's interesting because what it means is that one minus a third of i alpha plus a ninth of alpha squared is a special expression. It follows that one minus a third of i alpha minus a ninth of alpha squared is one plus omega plus omega squared where omega is minus a third of i alpha but more importantly where omega is a cube root of unity. So what we've got here has the form one plus omega plus omega squared where omega is a cube root of unity. Now that's interesting because you should remember that if omega is a complex cube root of unity then one plus omega plus omega squared will be zero. So therefore one minus a third of i alpha minus a ninth of alpha squared will be zero. Okay that's the end of that question. Let's move on to the next question I want to look at number five. This one tells us that z is cos theta plus i sin theta and then we have to use de Moivre's theorem to show that z to the power of n plus one over z to the power of n is equal to two times the cosine of n theta for any positive integer n. Now this is a really important question because this idea turns up a lot and you'll need to be able to prove this fact if you're asked to. Now z to the power of n is going to be cos theta plus i sin theta to the power of n and de Moivre's theorem tells us that that's cos n theta plus i sin n theta. Also one over z to the power of n which is z to the power of minus n is going to be cos theta plus i sin theta to the power of minus n. And again using de Moivre's theorem that must be cos of minus n theta plus i sin minus n theta which is cos n theta minus i sin n theta. And I hope now you can see that if we add up z to the power of n and one over z to the power of n the sin terms cancel out and the cos term doubles up. So z to the power of n plus one over z to the power of n is equal to two times the cosine of n theta. The next part of the question tells us to use this fact to express 16 cos to the power of 4 theta in the form something times cos 4 theta plus something times cos 2 theta plus something. Now we're going to use the thing we just proved. So remember z to the power of n plus one over z to the power of n is 2 cos n theta. Well how can we use this? Well one thing that you might note is that 16 times cos to the power of 4 theta is the fourth power of 2 cos theta. And 2 cos theta is z plus one over z. So what we've got here is the fourth power of z plus one over z according to the rule we just proved. Now let's raise this bracket to the power of 4. Using the binomial theorem we can expand these brackets and get z to the power of 4 plus 4z cubed times one over z plus 6z squared times the square of one over z plus 4z times the cube of one over z plus the fourth power of one over z. And that simplifies to z to the power of 4 plus 4z squared plus 6 plus 4 over z squared plus one over z to the power of 4. And if you reorder the terms here you'll get something really interesting. This is the same as z to the power of 4 plus one over z to the power of 4 plus 4 lots of z squared plus one over z squared plus 6. And you'll probably see here that z to the power of 4 plus one over z to the power of 4 is something we can use. And z squared plus one over z squared is something we can use. In fact, z to the power of 4 plus one over z to the power of 4 is 2 cos 4 theta according to the result from before. And z squared plus one over z squared is 2 cos 2 theta according to the result from before. So we've got 2 cos 4 theta plus 4 lots of 2 cos 2 theta plus 6. that's 2 cos 4 theta plus 8 cos 2 theta plus 6. And this is 16 cos to the power 4 theta expressed in the form something times cos 4 theta plus something times cos 2 theta plus something. Okay the next part of the question says given that cos 4 theta is 8 times cos to the power 4 theta, find two possible values of cos theta. We've got to give the answer in exact form. Okay so we've got the equation cos 4 theta is a half of 16 times cos to the power 4 theta. Now obviously we've just worked out an expression for the right-hand side. So it's equal to a half of 2 cos 4 theta plus 8 cos 2 theta plus 6. Or more simply cos 4 theta plus 4 cos 2 theta plus 3. Now if you look at the left and the right-hand sides here you'll see that the cos 4 theta's cancel out. So we're trying to find when 4 cos 2 theta plus 3 equals 0. The one strategy here would be to subtract 3 from both sides giving 4 cos 2 theta equals minus 3 and cos 2 theta equals minus 3 quarters. But this isn't getting us towards an exact solution. A better method is to go back slightly and to replace cos 2 theta with 2 cos squared theta minus 1. So now I've got the equation 4 lots of 2 cos squared theta minus 1 plus 3 equals 0. Or 8 cos squared theta minus 4 plus 3 equals 0. Or rather 8 cos squared theta minus 1 equals 0. 8 cos squared theta equals 1. Cos squared theta equals 1. And therefore cos theta is plus or minus 1 over 2 root 2. Okay that's the end of that question. Now let's move on to question 7. This tells us to show the 4 times sine cubed theta is identically equal to 3 sine theta minus sine 3 theta. Or we can prove this using De Moivre's theorem. According to De Moivre's theorem cos 3 theta plus i sine 3 theta is identically equal to the cube of cos theta plus i sine theta. And if you multiply out those brackets using the binomial theorem we get cos cubed theta plus 3 cos squared theta times i sine theta plus 3 cos theta times the square of i sine theta plus the cube of i sine theta. And that simplifies to cos cubed theta plus 3 i cos squared theta sine theta minus 3 cos theta sine squared theta minus i sine cubed theta. Or if we write it in terms of its real component and its imaginary component we've got the cos cubed theta minus 3 cos theta sine squared theta plus i times 3 cos squared theta sine theta minus sine cubed theta. So now if we equate the imaginary parts on the left hand side and the right hand side we discover that sine 3 theta is identically equal to 3 cos squared theta sine theta minus sine cubed theta. And now we're almost there. We simply have to replace cos squared theta with 1 minus sine squared theta giving us 3 lots of 1 minus sine squared theta times sine theta minus sine cubed theta. Multiply out those brackets to get 3 sine theta minus 3 sine cubed theta minus sine cubed theta and then 3 sine theta minus 4 sine cubed theta. And then we arrange and we've got that 4 sine cubed theta is identically equal to 3 sine theta minus sine 3 theta. Okay so this is a very important technique that you should remember involving De Moivre's theorem. The next part of the question says given that 4 cos cubed theta is identically equal to cos 3 theta plus 3 cos theta which by the way we could have worked out by equating the real parts from the previous equation. Anyway it tells us to use suitable compound angle formulae to show that 32 times sine cubed theta times cos cubed theta is identically equal to 3 sine theta minus sine 6 theta. Well you'll probably notice that 32 sine cubed theta cos cubed theta is 2 lots of 4 sine cubed theta times 4 cos cubed theta. And we know expressions for 4 sine cubed theta and 4 cos cubed theta. So it's natural to say that this is double 3 sine theta minus sine 3 theta times cos 3 theta plus 3 cos theta. And if we times out those brackets we get a complicated expression involving various sines and cosines. We've got 6 sine theta cos 3 theta plus 18 sine theta cos theta minus 2 sine 3 theta cos 3 theta minus 6 sine 3 theta cos theta. Let's write those in a different order because we're supposed to be able to use some compound angle formulae to simplify this. Let's write this as 18 sine theta cos theta minus 2 sine 3 theta cos 3 theta minus 6 sine 3 theta cos theta minus sine theta cos 3 theta. Now the point of that is that sine theta cos theta can be transformed into something to do with sine 2 theta. Sine 3 theta cos 3 theta can be transformed into something to do with sine 6 theta. And sine 3 theta cos theta minus sine theta cos 3 theta can be transformed into something to do with sine 3 theta minus theta. So we've got 9 knots of 2 sine theta cos theta minus 2 lots of sine 3 theta cos 3 theta minus 6 lots of sine 3 theta cos theta minus sine theta cos 3 theta and that would be the same as 9 times sine 2 theta minus sine 6 theta minus 6 lots of sine 3 theta minus theta using some suitable compound angle formulae. And that's just 9 sine 2 theta minus sine 6 theta minus 6 sine 2 theta which simplifies to 3 sine 2 theta minus sine 6 theta which is the result we were supposed to be proving. Before we go on note that there's actually a much simpler way of proving this ignoring the instruction in the question. We can also set about it by saying that 32 times sine cubed theta cos cubed theta is 4 times the cube of 2 sine theta cos theta but 2 sine theta cos theta is simply sine 2 theta so this is 4 times the cube of sine 2 theta. But we know from before that 4 times the cube of sine theta is 3 sine theta minus sine 3 theta. So therefore 4 times the cube of sine 2 theta must be 3 sine 2 theta minus sine 6 theta and this shows that 32 sine cubed theta cos cubed theta is 3 sine 2 theta minus sine 6 theta much more simply. Okay the last part of the question asks us to verify that the gradient of the curve with equation y equals sine cubed theta cos cubed theta at the point where theta equals pi by 4 is 0 and we have to find the nature of this stationary point. Okay well obviously we have to find dy by d theta. Okay well dy by d theta will be d by d theta of sine cubed theta times cos cubed theta and that'll be a nightmare to work out except that we know an expression for sine cubed theta cos cubed theta. We've got to find d by d theta of 1 over 32 3 sine 2 theta minus sine 6 theta. We can take the fraction outside the d by d theta so this is 1 over 32 times d by d theta 3 sine 2 theta minus sine 6 theta and that's 1 over 32 times 6 cos 2 theta minus 6 cos 6 theta. Now in theta is equal to pi over 4 this is giving us that dy by d theta is 1 over 32 times 6 cos pi by 2 minus 6 cos 3 pi by 2 which is simply 1 over 32 times 6 times 0 minus 6 times 0 which is obviously 0. So that proves that the gradient of the curve is 0 when theta equals pi by 4. To find the nature of the stationary point we need to differentiate again. So dy by d theta is 1 over 32 times 6 cos 2 theta minus 6 cos 6 theta and differentiating that to find d2y by d theta squared we've got d by d theta of 1 over 32 times 6 cos 2 theta minus 6 cos 6 theta which is 1 over 32 times dy by d theta 6 cos 2 theta minus 6 cos 6 theta which is 1 over 32 times minus 12 sine 2 theta plus 36 sine 6 theta and when theta equals pi over 4 that's going to be 1 over 32 times minus 12 sine pi over 2 plus 36 sine 3 pi over 2 which is 1 over 32 times minus 12 times 1 plus 36 times minus 1 which comes to minus one and a half. So d2y by d theta squared is negative when theta equals pi over 4 and therefore the stationary point is a maximum. Okay let's move on to question 10. This is asking us to find the Cartesian equation of the locus the modulus of z minus 2 minus 2i equals the square root of 2. Okay well if we write z in the form x plus i y we've got x plus i y minus 2 minus 2i equals the square root of 2 and therefore the modulus of x minus 2 plus y minus 2 times i equals the square root of 2. Well this means that the square of x minus 2 plus the square of y minus 2 is 2 so that's the Cartesian equation of this locus. We're now asked to find the Cartesian equation of the locus arg z minus 2 equals 3 pi over 4. Now this is going to be a half line which begins at 2 and makes an angle of 3 pi over 4 radians with the positive real axis and its gradient is obviously going to be minus 1. Now the full line would have equation y equals 2 minus x because it passes through both the x and y axes at 2. However we want a half line and the half line is just going to be where y equals 2 minus x and x is less than 2. Now we're told to use algebra to find the complex number which satisfies both low key. Well that means solving the simultaneous equations the square of x minus 2 plus the square of y minus 2 equals 2 along with y equals 2 minus x. We did that just by substituting y equals 2 minus x into the first equation giving us the square of x minus 2 plus the square of 2 minus x minus 2 equals 2. For more simply the square of x minus 2 plus the square of minus x equals 2, x square of minus 4 x square plus 4 plus x squared equals 2, and so on. 2 x squared minus 4 x squared plus 4 equals 2, 2 x squared minus 4 x squared plus 2 equals 0. You can just halve that eventually giving us that the square of x minus 1 is equal to 0, and that obviously only has one solution. It means that x is equal to 1. So that's the solution for x, x is 1 and that obviously means that y is also equal to 1 and therefore the complex number satisfying both equations therefore lying on both low key is 1 plus i. Now we're told to sketch the low key and state the geometrical relationship between them. Okay well first of all the half line given by y equals 2 minus x looks like this, and secondly the other locus is a circle with center 2 plus 2 i and radius the square root of 2. So that looks like this. Obviously the geometrical relationship between these is that the half line is a tangent to the circle. And by the way we could have told this from the fact that we only got one solution to the quadratic equation we were solving a minute ago. The fact that there's only one solution means that the tangent just touches the circle. If there were two solutions then the line would have cut the circle twice and if there were no solutions then the line wouldn't have met the circle at all. The fact that it meets the circle just once means that it just touches it and therefore that it's a tangent. Okay the next question is number 14. Here we're told that in an Argan diagram the point p represents the complex number z where the argument of z minus 3 over z plus i is pi by 4. First of all we're given two complex numbers minus 1 and 3i and we're asked to show that they're on the locus of p. Obviously that means showing that they satisfy this equation. So to begin with let's find what happens if we substitute minus 1 into the expression z minus 3 over z plus i. So we'll work out what minus 1 minus 3 over minus 1 plus i is actually equal to. Still equal to minus 4 over minus 1 plus i and we can simplify that by multiplying by minus 1 minus i over minus 1 minus i. That gives us 4 plus 4i over the square of minus 1 minus i squared which is 4 plus 4i over 1 minus minus 1 which is 4 plus 4i over 2 which is 2 plus 2i. And okay this is a complex number where the real and imaginary parts are the same so it lies on the line y equals x and its argument is clearly equal to pi over 4. Now let's look at the other complex number 3i and substitute that into the expression. Let's find 3i minus 3 over 3i plus i. Obviously that's 3i minus 3 over 4i and we can simplify that by multiplying by minus i over minus i. That gives us 3 plus 3i over 4 or 3 quarters plus 3 quarters of i. And again that's clearly on the line y equals x and so it's clearly got argument pi over 4. So that shows that both minus 1 and 3i lie on the locus that we're talking about. Okay now we're asked to sketch it and to label these points on the sketch. So it looks something like this. It's an arc of a circle which begins at 3 and goes anticlockwise round to the point minus i. And obviously it's a major arc. Minus 1 is there on the negative real axis and 3i is up there on the positive imaginary axis. So this is the sketch. The next part of the question tells us to find the center and radius of the circle of which the locus is a part. We've got to give the radius in exact form. Now we can do this just by looking at the sketch. Clearly the center of the circle is the same distance away from minus 1 as it is from 3 because they're both points on the circle. So that means that the center must be on the perpendicular bisector of the line joining minus 1 to 3 because that's the line of all the points which are the same distance from minus 1 as from 3. And by the way this is the line where x is equal to 1. In a similar way the center must be on the horizontal line that's the perpendicular bisector of the line joining minus i to 3i because obviously it's the same distance away from minus i as it is from 3i and that horizontal line is the set of all points which are the same distance away from 3i as minus i. And obviously this is the line y equals 1. Okay so that tells us where the center of the circle is. It's at x equals 1 y equals 1. In other words it's the complex number 1 plus i. To find the radius we just have to look at the distance between 1 plus i and the real number 3. And that distance is going to be the square root of 2 squared plus 1 squared because 2 is the distance between them along the real axis and 1 is the distance between them along the imaginary axis. So the radius is the square root of 5. And that's the end of that question. One more question, question 16. We told that p represents a complex number where z minus 2 equals z minus 2i. And first of all we've got to find the locus of p. Well this is easy. This is the locus of points which are the same distance away from 2 as they are from 2i because the modulus of z minus 2 is the distance of z from 2 and the modulus of z minus 2i is the distance of z from 2i. Well the locus of points which are the same distance away from 2 as from 2i is the perpendicular bisector of the line joining 2 to 2i. So it's a diagonal line like this which looks rather like y equals x. So now on to the meat of the question. This gives us a transformation from the z plane to the w plane defined by w equals z over z minus 2i. I've got to show that t maps the locus of p in the z plane to a circle in the w plane. I've also got to find a Cartesian equation of this circle. Okay so the method is to start with w equals z over z minus 2i and rewrite it to make z the subject. So first of all let's multiply both sides by z minus 2i. So w times z minus 2i equals z. Then multiply out the brackets so wz minus 2i w equals z, wz minus z equals 2i w and so on. We'll keep going factorizing that. w minus 1 times z equals 2i w and therefore z is 2i w over w minus 1. So now we've got z in terms of w. z is 2i w over w minus 1 and we know that the modulus of z minus 2 is the modulus of z minus 2i. So what we do now is we substitute our expression for z into the modulus equation and then we'll get an equation that only involves w. So we know that the modulus of 2i w over w minus 1 minus 2 is the modulus of 2i w over w minus 1 minus 2i. This is an equation that only involves w and if we simplify it it should be equivalent to the equation of a circle. So let's have a go at simplifying it. An obvious first step is to multiply everything by w minus 1 so that we don't have to deal with fractions and that gives us this. The modulus of 2i w minus 2 lots of w minus 1 is the modulus of 2i w minus 2i times w minus 1. Multiplying out the brackets we can simplify slightly especially on the right hand side which just turns into the modulus of 2i. Now it's a bit tricky having a complex number times w so let's factorize that out. So what we'll get is the modulus of 2i minus 2 times the modulus of w plus 2 over 2i minus 2 is equal to the modulus of 2i. Now the modulus of 2i minus 2 is 2 root 2 if you think about how far it is away from the origin and the modulus of 2i is just 2 and if you divide both sides by 2 times the square root of 2 we've now got the modulus of w plus 2 over 2i minus 2 is 1 over root 2. Now this is nearly simple enough to say what the locus of p is in the w plane. The only problem is that we don't know what complex number 2 over 2i minus 2 is so let's write that out. 2 over 2i minus 2 can be simplified by multiplying by 2i plus 2 over 2i plus 2. So that's 4i plus 4 over the square of 2i minus the square of 2 which is 4i plus 4 over minus 4 minus 4 or 4i plus 4 over minus 8 so that complex number is minus a half minus a half of i and now we know this we can say that the locus we're looking at is that the modulus of w minus a half minus a half of i is 1 over the square root of 2 or put a different way so that the modulus of w minus a half plus a half of i is 1 over the square root of 2 and this should be familiar this is the equation of a circle whose center is at a half plus a half of i and whose radius is 1 over root 2 and the Cartesian equation of that circle if we say that w is u plus iv is the square of u minus a half plus the square of v minus a half is equal to a half okay i hope you've found it useful having a look at these questions thank you very much for watching