The best of the best thing is you never left behind any option without explaining it. Be it the right or wrong option, you never fail to amazed in both the sides......... Very helpful... Thank you so much.
In convex set a line AB joining any two points A, B in the set lies completely within the set. Here f(H) is points inside unit circle except zero. If we connect two points in real line let one from negative real axis and other from +ve real axis ,when we connect that two point using a line definitely it will pass through zero but zero does not belong to f(H).There fore its not a convex set.
Please check dot product of (1,0) and (0,1)
Dot Product for the 1st Option is not correct it is an orthogonal actually
The best of the best thing is you never left behind any option without explaining it.
Be it the right or wrong option, you never fail to amazed in both the sides.........
Very helpful...
Thank you so much.
Sir please complete the ode in the same way. Please make videos on Wronkskian, Sturm livollis problem and other important topics
Thank you a lot sir🙏🌼🌼🌼🙏
It is real differentiable function that means imaginary part is zero
Sir complex analysis ke part-c ka bhi solution upload kar dijiye
Sir dot product of (1,0)and(0,1) is 0
Then take other example as f(z)=z^2
Ok sir.. thanku
@@DrHarishGargf(z)=z^2 still gives dot product 0
Too sir option konsa thik hoga
@DrHarishgarg sir , f(z)=z^2 also gives dot product zero
Y f(H) is not convex set
In convex set a line AB joining any two points A, B in the set lies completely within the set. Here f(H) is points inside unit circle except zero. If we connect two points in real line let one from negative real axis and other from +ve real axis ,when we connect that two point using a line definitely it will pass through zero but zero does not belong to f(H).There fore its not a convex set.