Complexo nos.

if 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

tell me the naswer dude,,,whatever u r getting...because i think book ans is wrong..thts why askd here...

2>

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

hmm 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

i 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