Question

if z-1/z+1 is purely imaginary number (z±-1),find the value of |z|

Answer 1

(z±√-1) like this ?

Answer 2

igtg, and idk this ........

Answer 3

pls describe this

Answer 4

Let \(z=a+bi\), then
\[\begin{align*}\frac{z-1}{z+1}&=\frac{(a-1)+bi}{(a+1)+bi}\cdot\frac{(a+1)-bi}{(a+1)-bi}\\&=\frac{a^2-1+b^2}{(a+1)^2+b^2}\\&=\frac{a^2+b^2-1}{a^2+2a+1+b^2}
\end{align*}\]
Notice that there are no factors of \(i\) here, which means we've reduced it to a real number. But \(\dfrac{z-1}{z+1}\) is purely imaginary, which means this number must be 0.
\[\frac{a^2+b^2-1}{a^2+2a+1+b^2}=0~~\Rightarrow~~a^2+b^2=1~~\Rightarrow~~|z|=\sqrt{a^2+b^2}=\sqrt1=1\]

Answer 5

@SithsAndGiggles I think there is mistake from numerator :) it should be a^2 +b^2+2bi-1

Answer 6

which one

Answer 7

@Loser66
Yes, you're right, I made a mistake with the computation, but it still works out nicely.

The point is, we want the real part of the numerator to be zero; I just happened to leave to leave out the imaginary part.

@ssw, the answer is still correct, there was just a slight computational error that does not change the end result.