It's not 1. Do you know the trick where powers of i repeat in a cycle of 4? i0=1,i1=i,i2=−1,i3=−i,i4=1,.... If you do 2015/4=503.75 and just focus on the stuff after the decimal point, you get 0.75. If you do 0.75*4 you get 3, so this tells us that [text]i^{2015}=i^3=-i. (I can do more examples of this if it's not clear)
It's not 1. Do you know the trick where powers of i repeat in a cycle of 4? i0=1,i1=i,i2=−1,i3=−i,i4=1,.... If you do 2015/4=503.75 and just focus on the stuff after the decimal point, you get 0.75. If you do 0.75*4 you get 3, so this tells us that [text]i^{2015}=i^3=-i
. (I can do more examples of this if it's not clear)
Do you write it in polar form z= sqrt2/2(cos45+isin45) and then z=(sqrt2/2)^1/4(cos45/4+isin45/4)
This is very close, but you need to remember that when we take the nth root of a complex number, there are exactly n distinct roots. So here, we expect 4 different numbers.
Do you use Euler's equation? e^ipi/4 = cos pi/4 + isin pi/4 Do you take the natural log of both sides to get rid of the e?
Close, all you need to do from here is work out in your calculator (or by memory!) what cos(pi/4) is and what sin(pi/4) is and tells you what z=x+iy are. (Because x=cos(pi/4), y=sin(pi/4) in this case).
If you want to show that p(z)=z3−z let's say has a root at z=1 then you just need to show that p(1)=13−1=0. Hence z=1 is a root. Apply this to your function with x=−i and show that if you plug that into your polynomial, you get 0 out.
If you want to show that p(z)=z3−z let's say has a root at z=1 then you just need to show that p(1)=13−1=0. Hence z=1 is a root. Apply this to your function with x=−i and show that if you plug that into your polynomial, you get 0 out.
This is very close, but you need to remember that when we take the nth root of a complex number, there are exactly n distinct roots. So here, we expect 4 different numbers.
I don't know what I'm doing wrong. I attached what I did for this question in the above post. Can you see what I did wrong?