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Complex number question helpppp

I can’t for the life of me figure out the second solution to part 2
Reply 1
Sorry I’m new to tsr so don’t know how to insert a photo. The question is on the image attached
Original post by hdconfused
Sorry I’m new to tsr so don’t know how to insert a photo. The question is on the image attached


For part (i) you found that z=Reiϕz = R e^{i \phi} where R>0R > 0 and π<ϕπ-\pi < \phi \leq \pi

Now you wish to determine the values of z\sqrt{z}, which by the above, is the same as z1/2=(Reiϕ)1/2=Reiϕ2z^{1/2} = (Re^{i \phi})^{1/2} = \sqrt{R}e^{i \frac{\phi}{2}}

So r=Rr = \sqrt{R} and θ=ϕ2\theta = \dfrac{\phi}{2} form a solution.

But there is one more root with an argument other than ϕ2\frac{\phi}{2}. See if you can find it.
(edited 5 years ago)
Reply 3
Original post by hdconfused
Sorry I’m new to tsr so don’t know how to insert a photo. The question is on the image attached

Okay, so you have it in the exponential form good, so to find roots in exponential form you a root to a nth power has nth solutions, and you know the angle repeats every 2pi, so set up 2 square roots with 2 exponential which are the same but have a different angle. Hopefully this helps?
For part 1 you need to put both the numerator and denominator in exponential form. So find the modulus and argument of each: numerator has mod 18 and arg pi/6 and denominator has mod 2 and arg -pi/6. So the fraction is 18e^(ipi/6) /2e^(-ipi/6). Simplify that using indices rules and you get 9e^(ipi/3).

Then for part 2 do the roots. (9e^(ipi/3))^(1/2). So again using basic indices laws you get 3e^(ipi/6). Now you've found one root then you basically need to add pi to the argument because the nth roots of a complex number on an argand diagram always form a regular polygon of order n. Then make sure your value for the argument meets the restriction, so you get the second root as 3e^(-5ipi/6).
(edited 5 years ago)
Reply 5
Original post by EleanorRosa
For part 1 you need to put both the numerator and denominator in exponential form. So find the modulus and argument of each: numerator has mod 18 and arg pi/6 and denominator has mod 2 and arg -pi/6. So the fraction is 18e^(ipi/6) /2e^(-ipi/6). Simplify that using indices rules and you get 9e^(ipi/3).

Then for part 2 do the roots. (9e^(ipi/3))^(1/2). So again using basic indices laws you get 3e^(ipi/6). Now you've found one root then you basically need to add pi to the argument because the nth roots of a complex number on an argand diagram always form a regular polygon of order n. Then make sure your value for the argument meets the restriction, so you get the second root as 3e^(-5ipi/6).


Thank you so much for the response but I still don’t understand where -5ipi/6 came from. I’m doing my a levels but we haven’t been taught ‘the nth roots of a complex number always form a regular polygon of order n’. I was able to figure out the other root through the basic laws of indices but I still don’t get how to get the other one
Reply 6
The Mark scheme says this. I still don’t get where the +-1/2pi i came from and then from there how’d it go to -5pi/6
Original post by hdconfused
The Mark scheme says this. I still don’t get where the +-1/2pi i came from and then from there how’d it go to -5pi/6


You need to be aware of the 'n polygon' theory here. I'll give you a quick rundown:

When we have a complex number zz on the plane, we can express it in terms of its modulus and argument in the form z=reiθz = re^{i \theta} where r=zr = |z| and θ=argz\theta = \arg z.

But now notice what happens if we keep rr constant, and slightly increase the argument. The complex number will start rotating about the origin anticlockwise. It will keep doing that the more we change it, and eventually it will come back to where it began. Precisely when we change the argument by 2π2\pi. This means if we add on an argument of 2π2\pi onto θ\theta, we get the exact same zz value.

This means we can write down zz equivalently as z=reiθ+2πiz = re^{i \theta + 2 \pi i}.

But we don't need to stop at 2π2 \pi. What if we rotate it once more by 2π2\pi? We will get back at the same position again! This means we can turn it as many times as we want by 2π2\pi and get back to the same zz.

Similarly, what if we decrease our argument instead of increasing?? This just means the complex number rotates CLOCKWISE about the origin instead, and the ideas remain fundamentally the same; we can decrease the argument by 2π2\pi and end up at the same place. Hence we can write z=reiθ2πiz = re^{i \theta - 2\pi i}. Again, by the same ideas, we can turn it backwards as many times as we want by 2π2\pi and end up at the same place.

This brings us to the catch here. If I denote 2πk2\pi k as some multiple of 2π2\pi, i.e. where k=0,±1,±2,k = 0 , \pm 1, \pm 2, \ldots, then I can add this to the argument of zz and our value would not change. This brings us to:

z=reiθ+2πkiz = re^{i \theta + 2\pi k i}.


In your question you are intersted in z\sqrt{z} which by this idea implies z=reiθ2+2πk2i\sqrt{z} = \sqrt{r} e^{i \frac{\theta}{2} + \frac{2\pi k}{2} i}.

Since r=9r = 9 and θ=π3\theta = \dfrac{\pi}{3} from part (i) we get that:

z=3eπ6i+πki\sqrt{z} = 3 e^{\frac{\pi}{6}i + \pi k i}

When k=0k = 0 we get the usual answer. When k=1k = -1 we get z=3eπ6iπi=3e56πi\sqrt{z} = 3e^{\frac{\pi}{6}i - \pi i} = 3 e^{-\frac{5}{6}\pi i}

I'm not sure where they are getting ±12πi\pm \dfrac{1}{2}\pi i from.
(edited 5 years ago)
Reply 8
Original post by RDKGames
You need to be aware of the 'n polygon' theory here. I'll give you a quick rundown:

When we have a complex number zz on the plane, we can express it in terms of its modulus and argument in the form z=reiθz = re^{i \theta} where r=zr = |z| and θ=argz\theta = \arg z.

But now notice what happens if we keep rr constant, and slightly increase the argument. The complex number will start rotating about the origin anticlockwise. It will keep doing that the more we change it, and eventually it will come back to where it began. Precisely when we change the argument by 2π2\pi. This means if we add on an argument of 2π2\pi onto θ\theta, we get the exact same zz value.

This means we can write down zz equivalently as z=reiθ+2πiz = re^{i \theta + 2 \pi i}.

But we don't need to stop at 2π2 \pi. What if we rotate it once more by 2π2\pi? We will get back at the same position again! This means we can turn it as many times as we want by 2π2\pi and get back to the same zz.

Similarly, what if we decrease our argument instead of increasing?? This just means the complex number rotates CLOCKWISE about the origin instead, and the ideas remain fundamentally the same; we can decrease the argument by 2π2\pi and end up at the same place. Hence we can write z=reiθ2πiz = re^{i \theta - 2\pi i}. Again, by the same ideas, we can turn it backwards as many times as we want by 2π2\pi and end up at the same place.

This brings us to the catch here. If I denote 2πk2\pi k as some multiple of 2π2\pi, i.e. where k=0,±1,±2,k = 0 , \pm 1, \pm 2, \ldots, then I can add this to the argument of zz and our value would not change. This brings us to:

z=reiθ+2πkiz = re^{i \theta + 2\pi k i}.


In your question you are intersted in z\sqrt{z} which by this idea implies z=reiθ2+2πk2i\sqrt{z} = \sqrt{r} e^{i \frac{\theta}{2} + \frac{2\pi k}{2} i}.

Since r=9r = 9 and θ=π3\theta = \dfrac{\pi}{3} from part (i) we get that:

z=3eπ6i+πki\sqrt{z} = 3 e^{\frac{\pi}{6}i + \pi k i}

When k=0k = 0 we get the usual answer. When k=1k = -1 we get z=3eπ6iπi=3e56πi\sqrt{z} = 3e^{\frac{\pi}{6}i - \pi i} = 3 e^{-\frac{5}{6}\pi i}

I'm not sure where they are getting ±12πi\pm \dfrac{1}{2}\pi i from.

Thank youuuuu!!!!! This was so so helpful and clear. Really appreciate it
Original post by hdconfused
Thank you so much for the response but I still don’t understand where -5ipi/6 came from. I’m doing my a levels but we haven’t been taught ‘the nth roots of a complex number always form a regular polygon of order n’. I was able to figure out the other root through the basic laws of indices but I still don’t get how to get the other one

Well, I don't know what A-Level syllabus you do, but mine (Edexcel) definitely has that idea in the syllabus. It helps if you set it up as an equation in the reverse (i.e. squaring to get z). So if arg(z^(1/2)) = n you know that 2n = k(2pi) + pi/3 where k is some integer. Solve that and n = kpi + pi/6. Now you can keep going indefinitely with this, but because of the restriction, you'll notice that once you get to 2pi + pi/6 the roots just repeat (maybe it heps to think about rotating them around an argand diagram?).

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