Math Problem Statement

\left\vert e^{-2z}\right\vert $$$$ <1

Solution

To solve the inequality e2z<1\left| e^{-2z} \right| < 1, let's analyze the expression step by step:

Step 1: Understand e2ze^{-2z}

The expression e2ze^{-2z} involves the exponential function with a complex argument z=x+iyz = x + iy, where xx and yy are the real and imaginary parts of zz, respectively. This can be expanded as: e2z=e2(x+iy)=e2xe2iy.e^{-2z} = e^{-2(x + iy)} = e^{-2x} e^{-2iy}.

Step 2: Find the modulus

The modulus of e2ze^{-2z} is: e2z=e2xe2iy=e2xe2iy.\left| e^{-2z} \right| = \left| e^{-2x} e^{-2iy} \right| = \left| e^{-2x} \right| \cdot \left| e^{-2iy} \right|. Since e2iy=1\left| e^{-2iy} \right| = 1 (as it's a point on the unit circle in the complex plane), the modulus simplifies to: e2z=e2x=e2x.\left| e^{-2z} \right| = \left| e^{-2x} \right| = e^{-2x}.

Step 3: Solve the inequality

The inequality e2z<1\left| e^{-2z} \right| < 1 thus becomes: e2x<1.e^{-2x} < 1.

Take the natural logarithm of both sides (since e2x>0e^{-2x} > 0): 2x<ln(1).-2x < \ln(1). 2x<0.-2x < 0. x>0.x > 0.

Final Solution

The condition for e2z<1\left| e^{-2z} \right| < 1 is: x>0.\boxed{x > 0}.

This means the real part of zz must be positive.

Would you like further explanation or assistance with a related problem? Here are 5 questions for deeper exploration:

  1. How does the imaginary part of zz affect e2z\left| e^{-2z} \right|?
  2. What happens if e2z>1\left| e^{-2z} \right| > 1?
  3. Can you explain the significance of the modulus for complex numbers?
  4. How would the inequality change if zz were purely real?
  5. What is the geometric interpretation of e2ze^{-2z} on the complex plane?

Tip: Remember that the modulus of a complex exponential depends only on the real part of the exponent.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Complex Numbers
Exponential Functions
Inequalities
Modulus of Complex Numbers

Formulas

Exponential function: \( e^{-2z} = e^{-2x}e^{-2iy} \)
Modulus of a complex number: \( \left| e^{-2z} \right| = \left| e^{-2x} \right| \cdot \left| e^{-2iy} \right| \)
Logarithmic inequality: \( e^{-2x} < 1 \) leads to \( x > 0 \)

Theorems

Properties of Exponential Functions
Logarithmic Properties
Modulus of Complex Numbers

Suitable Grade Level

Undergraduate Mathematics or Advanced High School (Grades 11-12)