Question

If u-v=(x-y)(x2
+4xy+y2)
and f(z)=u+iv is an analytic
function of z=x+iy , find f(z) is terms of z.

Answer 1

$f(z)=-i z^{3}+c_{1}$

Step-by-step explanation:

Here f(z) = u + iv…. (i)

if(z) = iu – v ………. (ii)

Adding (i) and (ii) we get

f(z) + if(z) = u + iv + iu – v

(1+ i) f(z) = (u – v)+ i( u + v)

Let(1+ i) f(z) = F(z), u – v = U and u + v = V, then

F(z) = U + iV

$\begin{aligned}&\text { If } U=u-v=(x-y)\left(x^{2}+4 x y+y^{2}\right)\\&\text { So, } \frac{\partial U}{\partial x}=x^{2}+4 x y+y^{2}+x-y(2 x+4 y)\\&=3 x^{2}+6 x y-3 y^{2}=\left(x_{1}, y_{1}\right)\\&\begin{array}{l}\text { And, } \frac{\partial U}{\partial x}=-\left(x^{2}+4 x y+y^{2}\right)+(x-y)(4 x+2 y) \\=3 x^{2}-6 x y-3 y^{2}=\left(x_{2}, y_{2}\right)\end{array}\end{aligned}$

Now (z1,0)= 3z2

and (z2,0)= 3z2

$\begin{aligned}&F(z)=\int\left[3 z^{2}-i\left(3 z^{2}\right)\right] d z+c\\&F(z)=(1-i) z^{3}+c\\&(1+i) f(z)=(1-i) z^{3}+c\\&f(z)=-i z^{3}+c_{1}\end{aligned}$

To learn more

1)Expand f(z)=Sin z in a taylor series about z=pi/4

https://brainly.in/question/6975046

Answer 2

Answer:

svsvvvvvsdsdv

Step-by-step explanation: