find the reflection of the complex no. 2-i in the straight line
z = iz
11this is the way i did it......
I ve taken z = iz as the line x+y=0 ths part i'm not sure whether i am right...
so the reflection of 2-i i.e (2,-1) in the line x+y=0 is coming out to be (1,-2) i.e 1-2i
please point out my mistake if am wrong somewhere...
1now given equation of straight line is y = -x
hence reflection of 2,-1 abt y= - x is 1 , -2
hence required complex no. is 1 - 2i
hence required answer [1]
1z(bar)=iz
is
y=x
reason
take
z = a+ia
z(bar)=a-ia
z=iz(bar) rotation by 90 degree
so it shud be
y=x
let reflection be
x,y
now we have
2+x=y-1
y=x+3
also we have
1+y=2-x
y=1-x
solve both
we get
x=-1 .y=2
so answer is
(-1,2i )
11but where am i wrong...plz see the graph that i used....whhy is it not y=-x....
plz tell me where i am going wrong.... [2]
1proof by graph
take
z=x-ix
now rotate it by 90 in anticlockwise direction
hence
iz= =z(bar)
11but RSP bhaiya is it necessary to take z=x-ix...why cant we take z=x+ix....
[12]
1i suppose it is done oral[3][3]........n answer i suppose shud come[1-2i]
1are u sure rpf.... wat u wanted to prove???[3][3].......LOL!!!!!
1@sandipan...........no doubt...u r a genius man[1]
@rpf........i knew it man....just saying was it really needed???[1]
