A: If z is a complex no. (z≠1) then \left|\frac{z}{\left|z \right|}-1 \right|\leq \left|argz \right|
R:In a unit circle chord AP≤ arc(AP)
62\left|e^{i\theta}-1 \right|=\left| cos\theta+isin\theta-1\right|=\left| 2sin^2 \frac{\theta}{2}+i2sin\frac{\theta}{2}cos\frac{\theta}{2}\right|=2sin \frac{\theta}{2} \text{Derivative of which is }cos{\frac{\theta}{2}} \text{hence the proof!}
1can anyone explain it one more time......a little graphically if possible
341I cannot see Nishant Sir's post, but just note that z\|z| lies on the unit circle. Lets say A corresponds to (1,0) and P is z\|z|. Then |
|z\|z| - 1| ≤ |arg z| expresses the fact that chord AP is less than the length of arc. [remember that -π≤ arg z ≤ π]
62this is an awesome way to look at it! :)
@prophet. it feels a bit humiliating when a person like you calls me sir :)
1thank u all geniuses.......
when solution comes the question seems too easy!!!
341@nishant sir - hey some of your posts are very good. So thats reason enough to call you sir :D
62but I genuinely believe that you have an edge over me in mathematics :)
