but can you also do 34 part ii?
following from that.but can you also do 34 part ii?
I gave your post some protectionSuppose y=0, then z+1/z=x+1/x=k
so |k|=|x+1/x|=|(x^2+1)/x|
we use the basic inequality of x^2+1>=2x (Based on (x-1)^2>=0)
Then |k| = |(x^2+1)/x| >= |(2x)/x|=2 so |k|>=2
Suppose x^2+y^2=1:
Then |x|<=1 (as x and y are real numbers)
so subbing (x^2+y^2=1) into the equation for k will give us:
k= x(1+1)/1= 2x
Since |x|<=1, then |k|= |2x|=2|x|<=2*1=2 so |k|<=2
I gave your post some protection
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