A "complex" problem

Algebra Level 2

Let z z be a complex number.

How many zeroes does e z e^z have on the complex plane?

0 1 An infinite number 2

Denton Young by Denton Young , 47, USA

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2 solutions

If z ∈ C then z = a + ib with a,b ∈ R ⇒ z \in \mathbb{C} \text{ then z = a + ib with a,b} \in \mathbb{R} \Rightarrow ∣ e z ∣ = ∣ e a + i b ∣ = ∣ e a ∣ ⋅ ∣ e i b ∣ = e a > 0 ⇒ e z = 0 has 0 solutions |e^{z}| = |e^{a + ib}| = |e^{a}| \cdot |e^{ib}| = e^{a} > 0 \Rightarrow e^z = 0 \text{ has 0 solutions} on the complex plane \text{ on the complex plane}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nicely done!

Denton Young - 5 years, 4 months ago

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thank you very much!

Guillermo Templado - 5 years, 4 months ago
Denton Young
Feb 10, 2016

In order for e z e^z to be zero, z z has to be negative infinity. Thus no zeroes exist on the complex plane.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

How do we know that "In order for e z e^z to be zero, z z has to be negative infinity"?

I agree that \lim_{ z \rightarrow { - \infty + 0 i } e ^ z = 0 . However, this doesn't explain why there cannot be any other solutions.

How do we know that "In order for e z e^z to be zero, z z has to be negative infinity"?

I agree that \lim_{ z \rightarrow { - \infty + 0 i } e ^ z = 0 . However, this doesn't explain why there cannot be any other solutions.

Calvin Lin Staff - 5 years, 3 months ago

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