Imaginary Numbers - Question Help

[Rahul]

I'm working my way through the Imaginary Numbers unit, and there were a few questions I was having trouble with. Since there are a few of them, I thought I might make a topic to get a little help from you guys about them. Thanks for the help, everyone!

a) If z and w are complex numbers such that |z+w|=|z-w|, prove that abs(argz-argw)=π/2.

When working on this question, I was able to prove that argz=arctan(a) and argw=arctan(-1/a). To prove the argument difference, I made the geometric observation that the reciprocal represents a reflection about y=0 and the multiplication by -1 represents a reflection about y=x; thus, geometrically, the transposed angle will differ from the original angle by π/2 rads. Is there a way to algebraically prove this? Or is this geometric proof valid and acceptable?

b) Prove that if z+1/z is real then either |z| or z is real.

I was able to prove the case z is real. However, for the case |z|, the furthest I got is that abs(|z|)=1. Is this valid enough to prove the case |z|?

c) F(x) is a polynomial of order 4 and has two real roots, x=1 and x=5. The graph of y=f(x) is tangent to the graph at x=5. Write in factorial form an expression for f(x) and hence sketch a possible graph of f(x).

It is stated that the expression for f(x) is f(x)=a(x-5)²(x-1)². However. I do not understand why this is the only possible expression… wait, got it. Only two real roots, x-5 must be squared cause derivative is zero, thus both factors are squared. Yay. But couldn’t it be x-5 as a square root, x-1 as a singular root, then one imaginary root?

Thanks for any help!

[flinquinnster]

Wow, this is good complex numbers revision

a) To be honest, I don't really understand your proof about the arguments being reciprocals of one another. I'm also not sure about the geometric observations you've made regarding these arguments - I would think that the difference between two numbers with arguments that are a and -a would be a reflection in the x-axis, and a reciprocal would also mean a reflection in the x-axis! I think that terminology might be an issue for me here, because I don't remember encountering the term 'reciprocal' in complex numbers. So I'm just going off this website here. From this, therefore, I can't really see how you've proved that the angle between z and w is pi/2. Perhaps I can't see it clearly now.

I took a very different approach to the question - it is still geometric, but quite different. I feel it is simpler, so hopefully it will come across. What I did was draw z and w on an Argand diagram, and then use parallelogram law to find the vectors z+w and z-w. As we know that |z+w| = |z-w|, this means that the length of the diagonals of this parallelogram are equal. As all quadrilaterals which have equal diagonals (rectangle, square) also have sides that are perpendicular to each other, this means that z and w, the sides of this quadrilateral, must be 90º apart. Hence, abs(argz-argw) = pi/2.

b) Well, it seems you've already answered your question - it asks you to show that either z or |z| real - so it implies that you only need to show one is applicable. However, I do think that it is definitely better to look and prove both.

z is real is fairly easy to prove, and you said that you got that already - I tried substituting at first with z = x+iy, but that was torture so I did a very simple realisation of 1/z and then that got me to the answer in about two lines.

I must say that this question is confusing me. Isn't |z| something that is always real? Isn't it the sum of the squares of the real and imaginary parts? As both x^2 and y^2 must be positive, surely |z| = sqrt(x^2 + y^2) must also be positive? If you've managed to do it (and I think I've sort of stumbled on this as well), then I think that showing abs|z| = 1 is sufficient to show that |z| must be real.

c) I think you're a bit confused here. The cardinal rule is that for polynomials with real co-efficients, which is generally assumed in most cases, then complex roots MUST occur in conjugate pairs. i.e. if 2+3i is a root, then 2-3i must also be a root. This means that the number of complex roots must be even. Hence, you usually can't have one complex root hanging by itself.

But yeah, you've got the right reasoning for why x=5 must be a double root - that is, I'm assuming, that the tangent of the graph at x = 5 is 0 (i.e. a stationary point).

Imaginary numbers is confusing - so if you do feel confused, preferably ask a teacher for clarification as well, because I can't guarantee that what I've written will make any more sense - in fact, complex numbers are giving me a bit of a headache right now

[Rahul]

Thanks for the help, flin! My thoughts:

a) I see your point, that's an interesting way of doing the question. I was working on an algebraic proof, however, and am sitting here right now using trigonometric identities: abs(arctanx-arccotx)=pi/2. I'm not sure how to proceed further, however - any ideas?

b) Wow, how did I not realize that? Obviously |z| is always real, thanks for pointing that out. Odd question, though, yeah, especially in the either wording.

c) My wording sucks. Yeup, I see why now. the real root theorem. Thanks!

Really appreciate it.