-
Want to level up your HSC prep on those harder Maths questions? Register now for the BoS Trials (7th October)!
complex number problems (1 Viewer)
- Thread starter lollol
- Start date
cheers
toadstooltown
1337 }{4><
- Joined
- May 1, 2005
- Messages
- 137
- Gender
- Undisclosed
- HSC
- N/A
It's been a long long time, but I'll take a stab
i) I'm going to assume p=pi, or some form of it, otherwise k will never be an integer.
[cis(4p/7)]^k=1
cis(4pk/7)=1
4pk/7=2npi
k=(7n.pi)/(2p) (I'm foolishly assuming p=pi =P)
k=7n/2 n=0,1,2,3...
lowsest n for k to be integter is 2, then k=7
ii)z^n=1
let w be a solution, so that w^n=1
w^nm=1^m
w^nm=1
(w^m)^n=1
so w^m is also a solution
ii)(z1+z2)/(z1-z2)=2i
z1+z2=2i.z1-2i.z2
z2(1+2i)=z1(2i-1)
z2=[z1(2i-1)]/(1+2i)
z2=z1×[(-1+2i)(1-2i)]/[(1+2i)(1-2i)]
z2=[z1×(3+4i)]/5
i) I'm going to assume p=pi, or some form of it, otherwise k will never be an integer.
[cis(4p/7)]^k=1
cis(4pk/7)=1
4pk/7=2npi
k=(7n.pi)/(2p) (I'm foolishly assuming p=pi =P)
k=7n/2 n=0,1,2,3...
lowsest n for k to be integter is 2, then k=7
ii)z^n=1
let w be a solution, so that w^n=1
w^nm=1^m
w^nm=1
(w^m)^n=1
so w^m is also a solution
ii)(z1+z2)/(z1-z2)=2i
z1+z2=2i.z1-2i.z2
z2(1+2i)=z1(2i-1)
z2=[z1(2i-1)]/(1+2i)
z2=z1×[(-1+2i)(1-2i)]/[(1+2i)(1-2i)]
z2=[z1×(3+4i)]/5