For z = x + iy, the magnitude |e^z| = e^x — it depends only on the real part. Here x = 3, so |e^(3+5i)| = e^3. The imaginary part 5 controls only the angle (argument) of the result, rotating the output 5 radians counterclockwise from the positive real axis, but leaving the distance from the origin unchanged. Option A conflates magnitude with adding exponent components; option C confuses real and imaginary parts; option D applies the magnitude formula for a + bi incorrectly to the exponent.
Question 2 Multiple Choice
Which statement correctly describes the periodicity of the complex exponential function?
Ae^z is periodic with period 2π — adding 2π to z gives the same output
Be^z is periodic with period 2πi — adding 2πi to z gives the same output
Ce^z is not periodic — it is entire and strictly increasing like the real exponential
De^z repeats with period π because sin and cos both have period π
The complex exponential has period 2πi (not 2π). Adding 2πi to z shifts the imaginary part by 2π, completing a full rotation: e^{z+2πi} = e^z · e^{2πi} = e^z · 1 = e^z. The key is that the period is purely imaginary — it lives in the vertical (imaginary) direction in the complex plane. Adding a real 2π does not return to the same value unless the imaginary part also wraps around. This periodicity is what makes the complex exponential non-injective, unlike the real exponential.
Question 3 True / False
The complex exponential e^z is injective (one-to-one): no two distinct values of z produce the same output.
TTrue
FFalse
Answer: False
The complex exponential is NOT injective. Because e^(z + 2πi) = e^z for all z, infinitely many inputs map to the same output — specifically, the entire vertical family {z + 2πki : k ∈ ℤ} all map to the same value. This is the fundamental difference from the real exponential, which is strictly increasing and injective. The fundamental domain 0 ≤ Im(z) < 2π is the largest strip where e^z is one-to-one, and this non-injectivity is exactly why the complex logarithm requires branch cuts.
Question 4 True / False
The complex exponential e^z avoids the value 0 — no complex number z satisfies e^z = 0.
TTrue
FFalse
Answer: True
For any z = x + iy, we have |e^z| = e^x > 0, since the real exponential e^x is strictly positive for all real x. Because the magnitude is always positive, e^z can never equal zero. This means the image of e^z is ℂ \ {0}: it covers every nonzero complex number (it is surjective onto ℂ \ {0}), but zero is permanently excluded. This fact becomes important in complex analysis — for example, it means e^z has no zeros, making 1/e^z = e^{-z} entire.
Question 5 Short Answer
Explain why the complex exponential is not injective, and describe the fundamental domain where it is one-to-one.
Think about your answer, then reveal below.
Model answer: The complex exponential is not injective because it is periodic with period 2πi: e^(z + 2πi) = e^z for all z, so the entire family {z + 2πki : k ∈ ℤ} maps to the same output. The fundamental domain where e^z is one-to-one is any horizontal strip of height 2π — conventionally the strip 0 ≤ Im(z) < 2π (or equivalently −π ≤ Im(z) < π). Every nonzero complex number has exactly one preimage in this strip.
The non-injectivity follows directly from the periodicity e^{2πi} = 1: adding 2πi to the exponent multiplies the output by 1, leaving it unchanged. This is the core reason the complex logarithm (the inverse of e^z) must be multi-valued and requires choosing a branch cut to make it single-valued.