Question

Can Milne-Thompson Method be used to find the Real part of f(z) when Imaginary part is given ??

Answer 1

I know it can be used to find the imaginary part when real part is given
by plugging in
x=z, y=0

but what about the other way round ?

Answer 2

If the imaginary part of the analytic function
\( W=f(z)\) is
\( V=x^2-y^2+x/(x^2+y^2 ) \)
find the real part U

Answer 3

Can't you just multiply your function f(z) by i to reverse the parts around?

Answer 4

Suppose you have f(z)=a+bi then that means i*f(z)=-b+ai and you can give it -b to find a? Honestly I don't know I've never heard of this thing before so I can't tell you, but that's just my guess.

Answer 5

I can try that...but I won't know for sure whether that works, as i don't have the answer....if anyone has solved such type of problem before....or anyone can give solved example from the internet, then that would be great

Answer 6

will use your method s the last resort...thanks :)

Answer 7

I'm not a mathematician (i.e. never heard of Milne-Thompson),
but it seems reasonable for
g(z)= i*f(z)
that i*Re(f(z)) = Im(g(z)) and -Im(f(z))= Re(g(z))

Answer 8

ok, thanks

Answer 9

\[ \frac{z-1}{iz + i } \]


\[ \text{z=1, } \frac{1-1}{i+ i } = 0 \\ \text{z=i, } \frac{i-1}{i^2 + i}=1 \\
\text{z=-1, } \frac{-1-1}{-i+ i }= \infty
\]

Answer 10

Put z=1 and w=0 in (A)
∴0=(a+b)/(c+d)
a=-b→(B)

Put z=i,w=1 in (A)
1=(ai+b)/(ci+d)
∴ai+b=ci+d
a=c,b=d →(C)

Put z=-1 ,w=∞ in (A)
∞=(-a+b)/(-c+d)
∴-c+d=0
c=d→(D)

so, a=1,b=-1, c=d=i

which means (B) and (D) are correct
not (C)

Answer 11

omg, i see a whole lot of ? in triangle boxes

Answer 12

refresh your browser

Answer 13

ohh!
since, a,b,c,d can be complex

i cannot conclude a=c, b=d from
ai+b=ci+d

Answer 14

yes, I used b= -a and d= c
and solved for c in terms of a
(ai - a)/(cz+c) = 1 when z= i

Answer 15

cool, thanks :)

Answer 16

I guess we have to remember the coefficients can be complex.

Answer 17

yep!