Logo
Pattern

Discover published sets by community

Explore tens of thousands of sets crafted by our community.

Maximum Modulus Principle

30

Flashcards

0/30

Still learning
StarStarStarStar

f(z) = 1/z on the annulus 1 < |z| < 2

StarStarStarStar

The maximum modulus is at |z| = 1, thus max|f(z)| = 1 since f(z) is not defined at z=0.

StarStarStarStar

f(z) = tan(z) in the rectangle with vertices at -pi/4, pi/4, pi/4 + pi*i, -pi/4 + pi*i

StarStarStarStar

The maximum modulus occurs at the boundary of the rectangle, with explicit calculation required due to the function's periodic nature.

StarStarStarStar

f(z) = sin(z) in the rectangle with vertices at -pi, pi, pi + pi*i, -pi + pi*i

StarStarStarStar

The maximum modulus occurs on the boundary of the rectangle and is |sin(pi)| = 0.

StarStarStarStar

f(z) = z in the circle |z| < 3 centered at the origin

StarStarStarStar

The maximum modulus occurs on the boundary at |z| = 3, thus max|f(z)| = 3.

StarStarStarStar

f(z) = Im(z^3) in the sector { z : |z| < 1, 0 ≤ Arg(z) ≤ π/4 }

StarStarStarStar

The function is not analytic, so the MMP does not apply. Im(z^3) is the imaginary part of z^3.

StarStarStarStar

f(z) = cos(z) in the strip defined by -2 < Re(z) < 2, -1 < Im(z) < 1

StarStarStarStar

The maximum modulus occurs on the boundary, but the domain is unbounded in the imaginary direction.

StarStarStarStar

f(z) = z^6 in the unit disk excluding the origin

StarStarStarStar

We cannot apply MMP since the domain is not closed due to the missing origin.

StarStarStarStar

f(z) = sin(pi*z) in the horizontal strip defined by -1 ≤ Im(z) ≤ 1 and -infinity < Re(z) < infinity

StarStarStarStar

The maximum modulus cannot be determined since the strip is unbounded in the real direction.

StarStarStarStar

f(z) = z^4 - 4z^2 + 4 in the unit disk

StarStarStarStar

The maximum modulus occurs on the boundary at |z| = 1, thus max|f(z)| = 1.

StarStarStarStar

f(z) = |z| in the entire complex plane

StarStarStarStar

The function is not analytic anywhere in the complex plane, so MMP does not apply.

StarStarStarStar

f(z) = z^2 in the unit disk

StarStarStarStar

The maximum modulus occurs on the boundary at |z| = 1, thus max|f(z)| = 1.

StarStarStarStar

f(z) = 1+2z+z^2 in the half plane Re(z) >= 0

StarStarStarStar

The maximum cannot be determined since the half-plane is unbounded and not a closed domain.

StarStarStarStar

f(z) = e^(1/z) in the disk |z| ≤ 2 excluding z=0

StarStarStarStar

The function is not analytic at z=0, so the domain is not closed at this point, and we cannot use MMP directly.

StarStarStarStar

f(z) = Log(z) in the set { z : 1/2 < |z| < 1, -pi/2 < Arg(z) < pi/2 }

StarStarStarStar

The maximum modulus occurs on the boundary given the function is analytic in the domain and it is max|Log(z)| = Log(1) = 0.

StarStarStarStar

f(z) = z/(z^2+1) in the square with vertices at 1+i, 1-i, -1-i, -1+i

StarStarStarStar

The maximum modulus occurs on the boundary of the square, but explicit calculation is required to find the maximum value.

StarStarStarStar

f(z) = (z-1)^2 in the half-disk { z : |z| < 1, Re(z) > 0 }

StarStarStarStar

The maximum modulus occurs on the boundary of the half-disk, but due to the half-disk shape, explicit calculation is needed.

StarStarStarStar

f(z) = z/(1-z) in the disk |z| < 1/2

StarStarStarStar

The maximum modulus occurs on the boundary at |z| = 1/2, thus max|f(z)| = 1 since 1-z ≠ 0 inside the disk.

StarStarStarStar

f(z) = z^3 in the disk |z-1| < 2

StarStarStarStar

The maximum modulus occurs on the boundary at |z-1| = 2, thus max|f(z)| = 2^3 = 8.

StarStarStarStar

f(z) = sin(z^2) in the domain bounded by the circle |z| = 2

StarStarStarStar

The maximum modulus occurs on the boundary at |z| = 2, thus max|f(z)| occurs there, but its exact value depends on the location on the boundary.

StarStarStarStar

f(z) = cosh(z) in the vertical strip defined by 0 ≤ Re(z) ≤ 1 and -infinity < Im(z) < infinity

StarStarStarStar

The maximum modulus cannot be determined since the strip is unbounded in the imaginary direction.

StarStarStarStar

f(z) = exp(z^2) in the set |z-i| < sqrt(2)

StarStarStarStar

The maximum modulus occurs on the boundary at |z-i| = sqrt(2), thus max|f(z)| occurs there, but its exact value depends on the location on the boundary.

StarStarStarStar

f(z) = 1/(z-1) in the unit disk minus the point z=1

StarStarStarStar

We cannot apply MMP as the domain is not closed due to the missing point z=1.

StarStarStarStar

f(z) = 1+2i*z in the ellipse with foci at -1 and 1 and sum of distances to foci equal to 4

StarStarStarStar

The maximum modulus occurs on the boundary of the ellipse, thus max|f(z)| occurs there, but its exact value requires computation.

StarStarStarStar

f(z) = e^z in the left half-plane

StarStarStarStar

The maximum cannot be determined since the left half-plane is unbounded and not a closed domain.

StarStarStarStar

f(z) = z^(-1/2) in the domain |z-1| < 1, z ≠ 1

StarStarStarStar

As the domain includes z = 0 where the function is not analytic, MMP does not apply.

StarStarStarStar

f(z) = exp(Re(z)) in the rectangle -1 ≤ Re(z) ≤ 1, -1 ≤ Im(z) ≤ 1

StarStarStarStar

The function is not analytic, so MMP does not apply. The modulus of f(z) increases with Re(z).

StarStarStarStar

f(z) = Re(z^3) in the disk |z+2i| ≤ 1

StarStarStarStar

The function is not analytic, so MMP does not apply. Re(z^3) is the real part of z^3.

StarStarStarStar

f(z) = sqrt(z) in the sector { z : |z| ≤ 1, 0 ≤ Arg(z) ≤ π/2 }

StarStarStarStar

The maximum modulus occurs on the boundary with Arg(z) = π/2 leading to max|f(z)| = sqrt(1) = 1.

StarStarStarStar

f(z) = (z+2)/(z^2+1) in the left half-disk defined by |z| < 1, Re(z) ≤ 0

StarStarStarStar

The maximum modulus occurs on the boundary of the half-disk, explicit calculations are necessary for the maximum value.

StarStarStarStar

f(z) = z + 1/z on the annulus 1/2 < |z| < 2

StarStarStarStar

The maximum modulus occurs on the boundary of the annulus, either at |z| = 1/2 or |z| = 2 depending on the function's behavior on these circles.

Know
0
Still learning
Click to flip
Know
0
Logo

© Hypatia.Tech. 2024 All rights reserved.