Ian H
Lv 7
Ian H asked in Science & MathematicsMathematics · 2 years ago

# COMPLEX UNSOLVED Suppose that f(z) is analytic, with f(0) = 3 - 2i and f(1) =6 - 5i . Find f(1 + i) if Im f'(z) = 6x(2y - 1)?

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• kb
Lv 7
2 years ago

Note that if f(z) = u(x,y) + i v(x,y), then f'(z) = u_x + i v_x.

Hence, we are given Im f'(z) = v_x = 6x(2y - 1) = 12xy - 6x.

Integrating with respect to x yields v(x,y) = 6x²y - 3x² + h(y) for some function h.

Next, we use the Cauchy-Riemann equations:

u_x = v_y ==> u_x = 6x² + h'(y)

u_y = -v_x ==> u_y = -12xy + 6x.

By the equality of mixed partial derivatives,

u_xy = u_yx ==> h''(y) = -12y + 6.

Integrating twice:

h'(y) = -6y² + 6y + A

==> h(y) = -2y³ + 3y² + Ay + B for some constants A and B.

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Hence, v(x,y) = 6x²y - 3x² - 2y³ + 3y² + Ay + B.

Returning to the Cauchy-Riemann equations:

u_x = v_y ==> u_x = 6x² - 6y² + 6y + A

u_y = -v_x ==> u_y = -12xy + 6x

Integrate each of these with respect to the appropriate variable:

u = 2x³ - 6xy² + 6xy + Ax + j(y)

u = -6xy² + 6xy + k(x) for some functions j and k.

Putting this all together, we obtain u(x,y) = 2x³ - 6xy² + 6xy + Ax + C.

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In summary:

f(z) = u(x,y) + i v(x,y)

......= (2x³ - 6xy² + 6xy + Ax + C) + i (6x²y - 3x² - 2y³ + 3y² + Ay + B).

Using f(0) = f(0 + 0i) = 3 - 2i, we find that 3 - 2i = C + Bi

==> B = -2 and C = 3.

Next, using f(1) = f(1 + 0i) = 6 - 5i, we find that

6 - 5i = (2 + A + C) + i (-3 + B) = (A + 5) - 5i

==> A + 5 = 6 and thus A = 1.

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Hence,

f(z) = (2x³ - 6xy² + 6xy + x + 3) + i (6x²y - 3x² - 2y³ + 3y² + y - 2).

Finally,

f(1 + i) = (2 - 6 + 6 + 1 + 3) + i (6 - 3 - 2 + 3 + 1 - 2) = 6 + 3i.

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Final remark:

We can rewrite f in terms of z by grouping terms of like total degree (in x and y).

f(z) = (2x³ + 6ix²y - 6xy² - 2iy³) + (-3ix² + 6xy + 3iy²) + (x + iy) + (3 - 2i)

......= 2(x³ + 3ix²y - 3xy² - iy³) - 3i(x² + 2ixy + y²) + (x + iy) + (3 - 2i)

......= 2(x + iy)³ - 3i(x + iy)² + (x + iy) + (3 - 2i)

......= 2z³ - 3iz² + z + (3 - 2i).

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I hope this helps!

• Ian H
Lv 7
2 years ago

My effort:

If f(z) = 2z^3 – 3z^2*i + constant

f'(z) = 6(z^2 – iz) = 6(x^2 – y^2 + y) + x(12y – 6)i

Im f'(z) = 6x(2y - 1)

When z = 0, f(0) = 3 - 2i, so,

f(z) = 2z^3 – 3z^2*i + 3 - 2i

f(1) = 2 – 3*i + 3 - 2i = 5 – 5i

Should be f(1) = 6 - 5i

How do I correct this ?