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complex numbers

complex numbers

by praveen upadhyay -
Number of replies: 1

hello sir and buddies i want  to know the solution of the following quetions as soon as possible..

1)the real part of sin(a+ib) is:

a)sina {(ex+e-x)/2}

b)cosa {(ex+e-x)/2}

c)sin a

d)none of this

2)if IzI=MAX{Iz-1I,Iz+1I} then

a)Iz+zI=1/2

b)z+z=1

c)Iz+zI=1

d)none of this

here z stands for conjugate of z. I I for mod

In reply to praveen upadhyay

Re: complex numbers

by Srinath Murthy -

sin(a+ib)=ei(a+ib)-e-i(a+ib)/(2i) which can be solved to find the real part.


When Re(z) > 0, IzI=Iz+1I

When Re(z) < 0, IzI=Iz-1I

In any case, Iz+zI=1