1State T/F:
Given :z1≠z2
1)Equation of circle through z1 and z2 as diameter:
|2z -z1 -z2|=|z1 - z2|
2)Equation of any circle through z1 and z2:
|2z -z1 -z2|-|z1 - z2| +k Real( (z-z1)/(z-z2) ) =0
whrer k is a constant.
9 Answers
62X= (1+cosx+isinx/1+cosx-isinx)
1+cosx= 2cos2x/2
X= 2cos2x/2 + 2isinx/2cosx/2
2cos2x/2 - 2isinx/2cosx/2
= eix
11If Re(z)>0
then |1+z+z2.......+zn|=
The ans should be in terms of |z| and n
1if |z|≠0,1 ,then all circles passing through z and 1/z are:
a) concentric with |z|=1
b)co-axal
c)orthogonal with |z| =1
62|z-2| = |z-1|
Then draw the Diagram in the argand plane..
it is the set of points equidistant from (2,0) and (1,0) which is x=1.5
Re(Z)=1.5
I dont see how you can eliminate 3|z|2
The answer should be cannot be determined...
62|2z-i/ 5z+1|=m is the eqn of a straight line if m = ???
|2z-i|= |5z+1|.m
|z-i/2|= |z+1/5|.5m/2
For straight line, 5m/2 has to be equal to 1
m=2/5
62State T/F:
Given :z1≠z2
1)Equation of circle through z1 and z2 as diameter:
|2z -z1 -z2|=|z1 - z2|
2)Equation of any circle through z1 and z2:
|2z -z1 -z2|-|z1 - z2| +k Real( (z-z1)/(z-z2) ) =0
whrer k is a constant.
First equation is true... (That is bcos distance from the fixed point (z1+z2)/2 is constant.
second Equation is a circle.. and it passes through z1 the only problem in the form in which this equation is written is that the function gets undefined at z2! It will not satisfy z=z2 ... hence 2 is false!
