24tell me the naswer dude,,,whatever u r getting...because i think book ans is wrong..thts why askd here...
Q1 Dividing f(z) by z-i gives remainder i
and diviiding by z+i we get remainder 1+i
If f(z) is diivded by z2+1,ten remainder is ??
Q2 Represent on argand plane
lz+il<lz-xl<lz-il
Q3 If lzl≤1 and lωl≤1 then prove lz-ωl2≤(lzl-lωl)2+(argz-argω)2
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6 Answers
1if f(x) is divided by x-m
then remainder ll b given by
f(m)
bcoz one can write
f(x)=(x-m)*something + remainder
i.e
f(m)= remainder
when f(z) is divided by z^2 +1
we have
f(z) =(z^2+1)*something + remainder
i.e
of form f(z) =(z^2+1)P+ R
here R can be of degree 1 i.e
R=az+b
so f(z)=(z^2+1)P+ az+b
now use the conditions given :
we get
f(i)=ai +b=i
f(-i) =-ai +b = 1+i
we can get a and b from these two conditions
and hence the remainder
2412>
z= x+iy
z-x= iy
|z-x| =1
z-i=x + i(y-1)
|z-i| =√x^2+(y-1)^2
z+i = x +i(y+1)
|z+i| =√x^2 +(y+1)^2
**
|z+i|<|z-x|
x^2 +(y+1)^2 <1 interior of the circular region represented
**
lz-xl<lz-il
1<x^2 + (y-1)^2
exterior of the circular region represented
take the common portion of both the regions represented
1is the answer to first one
1/2 + i/2(z+2)
or in the desired form
iz/2 + [1/2 + i]
1hmm third one did take tym
3>
let z=ae^iα
w= be^iβ
|z-w|2=a2+b2-2abcos(α-β)
=a2+b2 -2ab +2ab[1 - cos(α-β)]
=(a-b)2 + 2ab * 2sin2 [(α-β)/2]
=(a-b)2 + 4absin2 [(α-β)/2]
now
a<=1 and b<= 1 as given
so ab<=1
still thinking ahead...w8
1i guess if we use
limits concept ahead for small angles
sinx =x
v ll get the answer
but hw can one say that the angle [α- β] is small to hav this approximation