STEP Support Programme. STEP 2 Complex Numbers: Solutions

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Sydney Singleton
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1 STEP Support Programme STEP Complex Numbers: Solutions i Rewriting the given relationship gives arg = arg arg = α. We can then draw a picture as below: The loci is therefore a section of the circle between = 0 and =. Since 0 < α π the section will be above the real axis, and it will be the edge of the the major segment of the circle. STEP Complex Numbers: Solutions

2 Alternatively Let = x + iy. We then have: arg = α arg arg = α argx + iy argx + iy = α [ ] argx + iy argx + iy = α argx + iy argx + iy = α + argx + iy argx + iy y x y x + y x y = α x xy yx xx + y = α y x x + y = α If we let k = α, then this becomes the equation x x + y ky = 0 i.e. x + y k = k + 4 Which is the equation of a circle. Bear in mind that you would still need to justify which part of the circle is actually the locus of. You can do this by drawing a diagram, as before, or noting that if arg = α and 0 < α π then the imaginary part of must be positive. Considering gives us: x + iy = x + iy [x + iy][x iy] = x + y = xx + y + ixy x y x + y = x + y x x + y y + i x + y And so for the imaginary part to be positive we need y > 0, i.e. must lie above the real axis. STEP Complex Numbers: Solutions

3 ii If 0 < α π then π < α π π. In this case we have a section of circle below the real axis, and the locus will be the edge of a minor segment of the circle, like the diagram below: As before, you could use an algebraic approach if you prefer. iii We have: = = = This means that the locus is all the points where the disce of from the origin is the same as the disce of from the point, 0, i.e. it is the perpendicular bisector of the points 0, 0 and, 0. This will be the vertical line x = 0.5. STEP Complex Numbers: Solutions 3

4 Alternatively Let = x + iy. We have: = x + iy x + iy = x x + y + ixy yx x + y = x x + y + iy x + y = x + y x + y x + y = x + y x + y = x + y x + y x + y = x + y x + y + x x x + y + y = x + y x + y = x x + y So either x + y = 0, which is not possible as this would imply that = 0, or x = which gives the line x =. Although I have not done it here, the question did ask you to sketch the locus, so you would need to include a sketch! For the last part, we have: arg w = arg + arg 3 4 arg 4 arg 3 Draw a diagram which might look like this: STEP Complex Numbers: Solutions 4

Then: arg = α arg 3 4 = β arg 4 = γ arg 3 = δ and so arg + arg 3 4 arg 4 arg 3 = α + β + γ δ Add on some extra lines and angles using the fact that alternate angles are equal and we have: We now have

5 Then: arg = α arg 3 4 = β arg 4 = γ arg 3 = δ and so arg + arg 3 4 arg 4 arg 3 = α + β + γ δ Add on some extra lines and angles using the fact that alternate angles are equal and we have: We now have the equations: α + x + δ = π β + γ y = π x + y = π 3 Then gives us: α + δ β γ + x + y = 0 and using 3 we can write this as: α + β + γ δ = π Hence arg w = π, and so w is on the real axis and is real!. STEP Complex Numbers: Solutions 5

6 Expanding the brackets gives us: cos + i sin cos φ + i sin φ = cos cos φ sin sin φ + i cos sin φ + sin cos φ = cos + φ + i sin + φ * We know that the statement cos + i sin n = cos n + i sin n is true when n =. Assume it is true when n = k, so we have cos + i sin k = cos k + i sin k. Now consider the case when n = k +. cos + i sin k+ = cos + i sin k cos + i sin = cos k + i sin kcos + i sin = cos k + + i sin k + = cosk + + i sink + where the second line uses the n = k case and the third line uses with φ = k. Hence if it is true for n = k then it is true for n = k +, and as it is true for n = it is true for all positive integers n. Expanding 5 i + i gives us: 5 i + i = 4 0i + i = i Using what we know about arguments we have arg5 i + arg + i = arg34 + 4i. Calculating the angles and using a diagram to ensure we get the signs correct gives us: 5 + = 4 34 π 5 + as required = 7 7 = π 4 For the last part, you need to work out what complex numbers to use. Comparison with before suggests that it might be worth considering π Consider + i0 i4 i 3 by looking at the arguments in the required result. We have: + i0 i4 i 3 = + 9i 64 48i + i = + 9i 5 47i = i = i So we have + i0 i4 i 3 = i. Taking arguments gives us π 4 3 = and hence we have: as required = π There are different approaches to this question, using slightly different complex numbers such as 0 + i. STEP Complex Numbers: Solutions 6

7 3 Substituting into w gives us: Therefore we have u = + ix + iy w = i + x + iy = y + ix x + i + y y + ix x i + y = x + i + y x i + y = x y + x + y + i x y + y x + + y = x + i x + y x + + y x x + + y and v = x + y x + + y. This part of the question is called the stem of the question. The results that you have shown here can be assumed now throughout the rest of the question. Note that this only holds for stem results, you cannot assume that a result shown in part i holds in part ii etc. i Setting x = / and y = 0 gives u = + cos = + cos = sin cos sin + cos = sin = sin and v = + cos = + cos = sin cos sin + cos = cos This means that we have u + v = sin + cos =. However, φ is undefined for φ = π, so we need to exclude = π, i.e. the point sin π, cos π = 0, is excluded. Setting x = / means that taking values of in the range π < < π will give x values in the range < x <. STEP Complex Numbers: Solutions 7

8 ii This case is almost identical to the previous one except that x is restricted to < x <. This means we need to restrict so that < < i.e. π < < π. Since u = sin we have < u < and since v = cos we have < v < 0. Hence the locus is the semi-circle which is part of u + v = which lies strictly below the v-axis. iii If we let x = 0 we get u = 0 and v = y + y = y using the results from the y + stem of the question. Hence we are restricted to the v-axis and as < y < we have < v < 0. Therefore the locus is the negative v-axis i.e. the negative imaginary axis. iv As this looks similar to part i we can try using x = again. This gives: u = + 4 This doesn t look promising as the denominator will not simplify nicely. Instead try using x =, which gives: u = = + cos = + cos = sin cos = sin We also get: v = cos = + cos = sin cos = where the last step uses cos A = sin A. Like in part i we cannot have = π, so the point sin π, cos π = 0, is excluded. STEP Complex Numbers: Solutions 8

9 The final stage is to describe this locus fully. As we have sin and cos involved it is likely to be something circle related. We have: sin + cos = u + v = 4u + 4 v = u + v = 4 u + v = 4 hence the locus is a circle radius, centre 0, with the point 0, excluded. I haven t included any sketches in this solution, but a very good way of making your answers clear would be to sketch the loci make sure the labels on your axes are correct, i.e. u and v on the real and imaginary axes. You can use empty circles to show the excluded points. STEP Complex Numbers: Solutions 9

4 Using = gives: a c = a ca c = aa + cc ca ac It is helpful to draw a diagram to show the points in the argand plane representing a and c: Then the triangle has a right angle at A if and only if c =

10 4 Using = gives: a c = a ca c = aa + cc ca ac It is helpful to draw a diagram to show the points in the argand plane representing a and c: Then the triangle has a right angle at A if and only if c = a + a c, i.e.: cc = aa + aa + cc ca ac 0 = aa ca ac aa = ca + ac The circle and gent could look something like: The equation of the circle will be c = a c. This can be written as: c c = a ca c + cc c c = aa + cc ca ac STEP Complex Numbers: Solutions 0

11 Since the gent is perpendicular to the circle we have aa = ca +ac or aa ca ac = 0 and so we can write the equation of the circle as c c + aa = 0. P will lie on this circle if and only if: abab abc cab + aa = 0 aba b abc ca b + aa = 0 i.e. And P will lie on this circle if and only if: a b a b a b c c a b + aa = 0 i.e. aa bb ac b ca b + aa = 0 multiply by bb 0 to give aa abc ca b + aa bb = 0 This last line is the exactly same condition as, so point P is on the circle if and only if point P is on the circle. For the converse we cannot assume that OA is a gent, so we cannot use the condition aa = ac + ca. Going back to c = a c gives: c c = a a + cc c c = aa + cc ca ac If both P and P lie on this circle then we have: abab cab abc = aa ca ac and 4 a a a a c b b b b c = aa ca ac i.e. 5 aa ca b abc = aa ca ac bb 6 If we can show that aa ca ac = 0 then we will have shown that OAC = 90 and hence that OA is perpendicular to AC and so the line OA is a gent to the circle. If we take equation 4 and add aa to both sides, and take equation 6 and add aa bb we get the equations: aa bb ca b abc + aa = aa ca ac and 7 aa ca b abc + aa bb = aa ca ac bb 8 The left hand side of equations 7 and 8 are the same, so if we consider 8 7 we get: 0 = aa ca ac bb aa ca ac 0 = aa ca ac bb and since we are told that bb we have aa ca ac = 0 and the line OA is a gent to the circle. Note that if bb = then b = b. This means that a b = ab and hence P = P. STEP Complex Numbers: Solutions

STEP Support Programme. STEP III Complex Numbers: Solutions

STEP Support Programme. STEP III Complex Numbers: Solutions STEP Support Programme STEP III Complex Numbers: Solutions 1 A primitive nth root of unity is one that will generate all the other roots. This means that a is a primitive root of unity iff 1 a, a, a,...,