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FP2 complex numbers problem

The z-plane is mapped onto the w-plane by the transformation w=z+1/z . Given that z lies on the circle |z|=1, show that w lies on an interval of the real axis. Identify this interval precisely.

I've done:
w=(z^2 +1 )/z
|w|=|z^2 +1|/|z|
|w|=|z^2 +1|

Not sure what to do next
Original post by bobbricks
The z-plane is mapped onto the w-plane by the transformation w=z+1/z . Given that z lies on the circle |z|=1, show that w lies on an interval of the real axis. Identify this interval precisely.

I've done:
w=(z^2 +1 )/z
|w|=|z^2 +1|/|z|
|w|=|z^2 +1|

Not sure what to do next


I'd take another approach.

1z=1x+iy=xiyx2+y2=?\dfrac{1}{z}=\dfrac{1}{x+iy}= \dfrac{x-iy}{x^2+y^2}=?
Reply 2
Original post by BuryMathsTutor
I'd take another approach.

1z=1x+iy=xiyx2+y2=?\dfrac{1}{z}=\dfrac{1}{x+iy}= \dfrac{x-iy}{x^2+y^2}=?


So 1/z =x-iy as |z|=x^2 + y^2 =1 ?

So |w|=x-iy+1 ...?
Original post by bobbricks
So 1/z =x-iy as |z|=x^2 + y^2 =1 ?

So |w|=x-iy+1 ...?



Yes, 1/z=x-iy.

So what is z+1zz +\dfrac{1}{z}?
Reply 4
Original post by BuryMathsTutor
Yes, 1/z=x-iy.

So what is z+1zz +\dfrac{1}{z}?


Z+1/Z = 2x

so |w|=2|x| ...?
Original post by bobbricks
Z+1/Z = 2x

so |w|=2|x| ...?



and 1x1-1 \le x \le 1.
Reply 6
Original post by BuryMathsTutor
and 1x1-1 \le x \le 1.


How do you know that? :s-smilie:
Original post by bobbricks
How do you know that? :s-smilie:


Your question said: "Given that z lies on the circle |z|=1".
Reply 8
Original post by bobbricks
The z-plane is mapped onto the w-plane by the transformation w=z+1/z . Given that z lies on the circle |z|=1, show that w lies on an interval of the real axis. Identify this interval precisely.

I've done:
w=(z^2 +1 )/z
|w|=|z^2 +1|/|z|
|w|=|z^2 +1|

Not sure what to do next



although this particular question can be done in Cartesian
this type of problem is solved by the substitution z = e, which represent a unit circle

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