Question

find the analytic function whose real part is u = x³ - 3xy+ 3x² - 3y² +1​

Answer 1

Given: real part u=x3−3xy2+3x2−3y2+1

Let ϕ1(x,y)=∂u∂x,ϕ2(x,y)=∂u∂y

i.e.

ux=3x2−3y2+6x=ϕ1(x,y) and uy=−6xy−6y=ϕ2(x,y)

By Milne Thompson

f1(z)=ϕ1(z,0)−iϕ2(z,0)

i.e. substitute, x = z, y = 0 in ϕ1 & ϕ2

∴ϕ1(z,0)=3z2+6z

ϕ2(z,0)=0

∴f1(z)=3z2+6z−i(0)=3z(z+2)

Integrating the above equation

f(z)=∫3z2+6zdz=3z33+6z22+c=z3+3z2+c

Now, Z = x + iy

f(z)=(x+iy)3+(x+iy)2+cf(z)=x3+3x2(iy)+3x(iy)2+(iy)3+3(x2+2xyi+(iy)2)∴u+iv=(x3+3x2−3xy2−3y2)+i(3x2y−y3+6xy)∴u=(x3+3x2−3xy2−3y2) & V=(3x2y−y3+6xy)

Answer 2

Answer:

The analytic function is $z^{3}+3z^{2}+i(\frac{3z^{2}}{2} )$

Step-by-step explanation:

Given that,

The real part of the function is $u=x^{3}-3xy+3x^{2}-3y^{2}+1$

To find the analytical function we shall find the partial derivatives of above function with respect to x and y and then find the value of those derivatives at any arbitrary point on z-axis.

The partial derivative of above function with respect to x is

$\frac{\partial u}{\partial x} =3x^{2}-3y+6x$

Substituting x=z,y=0 in above expression we get,$f_{1}(z)=3z^{2}+6z$

Similarly,partial derivative of above function with respect to y is

$\frac{\partial u}{\partial y} =-3x-6y$ and substituting x=z,y=0 we get

$f_{2}(z)=-3z$

Now the analytical function is given by the relation

$f(z)=\int f_{1}(z)dz-i\int f_{2}(z)dz+c$

Substituting the deduced functions in above relation,we get

$f(z)=\int(3z^{2}+6z)dz-i\int(-3z)dz+c\\=z^{3}+3z^{2}+i(\frac{3z^{2}}{2} )$

Therefore,the analytic function is $z^{3}+3z^{2}+i(\frac{3z^{2}}{2} )$

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