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).
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
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 z on the plane, we can express it in terms of its modulus and argument in the form z=reiθ where r=∣z∣ and θ=argz.
But now notice what happens if we keep r 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 2π. This means if we add on an argument of 2π onto θ, we get the exact same z value.
This means we can write down z equivalently as z=reiθ+2πi.
But we don't need to stop at 2π. What if we rotate it once more by 2π? We will get back at the same position again! This means we can turn it as many times as we want by 2π and get back to the same z.
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 2π and end up at the same place. Hence we can write z=reiθ−2πi. Again, by the same ideas, we can turn it backwards as many times as we want by 2π and end up at the same place.
This brings us to the catch here. If I denote 2πk as some multiple of 2π, i.e. where k=0,±1,±2,…, then I can add this to the argument of z and our value would not change. This brings us to:
z=reiθ+2πki.
In your question you are intersted in z which by this idea implies z=rei2θ+22πki.
Since r=9 and θ=3π from part (i) we get that:
z=3e6πi+πki
When k=0 we get the usual answer. When k=−1 we get z=3e6πi−πi=3e−65πi
You need to be aware of the 'n polygon' theory here. I'll give you a quick rundown:
When we have a complex number z on the plane, we can express it in terms of its modulus and argument in the form z=reiθ where r=∣z∣ and θ=argz.
But now notice what happens if we keep r 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 2π. This means if we add on an argument of 2π onto θ, we get the exact same z value.
This means we can write down z equivalently as z=reiθ+2πi.
But we don't need to stop at 2π. What if we rotate it once more by 2π? We will get back at the same position again! This means we can turn it as many times as we want by 2π and get back to the same z.
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 2π and end up at the same place. Hence we can write z=reiθ−2πi. Again, by the same ideas, we can turn it backwards as many times as we want by 2π and end up at the same place.
This brings us to the catch here. If I denote 2πk as some multiple of 2π, i.e. where k=0,±1,±2,…, then I can add this to the argument of z and our value would not change. This brings us to:
z=reiθ+2πki.
In your question you are intersted in z which by this idea implies z=rei2θ+22πki.
Since r=9 and θ=3π from part (i) we get that:
z=3e6πi+πki
When k=0 we get the usual answer. When k=−1 we get z=3e6πi−πi=3e−65πi
I'm not sure where they are getting ±21πi from.
Thank youuuuu!!!!! This was so so helpful and clear. Really appreciate it
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?).