Seems true

Algebra Level 4

True or False?

Let α \alpha , β \beta be complex numbers with non-positive real parts. Then ∣ e α − e β ∣ ≤ ∣ α − β ∣ \large |e^{\alpha} - e^{\beta}| \le |\alpha - \beta|

Notation: e ≈ 2.718 e \approx 2.718 denotes the Euler's number .

True False

Ravneet Singh by Ravneet Singh , 30, India

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1 solution

Alex Burgess
May 2, 2019

Let α = x + i y , β = u + i v \alpha = x+iy, \beta = u+iv with x − u = z ≥ 0 , ∣ y − v ∣ = θ x-u = z \geq 0, \lvert y - v \rvert = \theta .

∣ e α − e β ∣ = ∣ e x e i y − e u e i v ∣ = e x ∣ e i y − e − z e i v ∣ \lvert e^{\alpha} - e^{\beta} \rvert = \lvert e^x e^{iy} - e^u e^{iv} \rvert = e^x\lvert e^{iy} - e^{-z}e^{iv} \rvert .

Hence,

∣ e α − e β ∣ = e x ∣ 1 − e − z e i θ ∣ = e x ∣ ( 1 − e − z ( cos ⁡ θ + i sin ⁡ θ ) ∣ = e x ( 1 + e − 2 z − 2 e − z cos ⁡ θ ) \lvert e^{\alpha} - e^{\beta} \rvert = e^x\lvert 1 - e^{-z}e^{i\theta} \rvert = e^x\lvert (1 - e^{-z}(\cos{\theta} + i\sin{\theta}) \rvert = e^x(1+e^{-2z} -2e^{-z}\cos{\theta} ) .

cos ⁡ x ≥ 1 − x 2 2 \cos{x} \geq 1 - \frac{x^2}{2} and 1 − e − x = x − x 2 2 + x 3 6 + . . . ≤ x 1-e^{-x} = x-\frac{x^2}{2}+\frac{x^3}{6} + ... \leq x , therefore:

∣ e α − e β ∣ ≤ e x ( 1 − e − z ) 2 + e y θ 2 ≤ z 2 + θ 2 = ∣ α − β ∣ \lvert e^{\alpha} - e^{\beta} \rvert \leq e^x(1-e^{-z})^2 + e^y\theta^2 \leq z^2 + \theta^2 = {\lvert \alpha - \beta \rvert} . []

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Hello, Sir. I only understand your solution up to the dividing line, but I noticed that you have a typo. You are missing a square root in that last expression (just before the dividing line). I'm afraid this little blunder might affect the rest of your solution.

James Wilson - 5 months ago

I understand your solution now without the mistakes. You are very smart. Here is what I think it should say without the typos. Let the variables be defined in the same way as you (Alex Burgess) define your variables, except let θ = v − y \theta = v-y .

∣ e α − e β ∣ 2 = ∣ e x e i y − e u e i v ∣ 2 = e 2 x ∣ 1 − e − z e i θ ∣ 2 = e 2 x ∣ 1 − e − z ( cos ⁡ θ + i sin ⁡ θ ) ∣ 2 |e^\alpha-e^\beta|^2 = |e^xe^{iy}-e^ue^{iv}|^2=e^{2x}|1-e^{-z}e^{i\theta}|^2=e^{2x}|1-e^{-z}(\cos{\theta}+i\sin{\theta})|^2

= e 2 x ( 1 + e − 2 z − 2 e − z cos ⁡ θ ) = e 2 x ( 1 + e − 2 z + ( − 2 + 2 − 2 cos ⁡ θ ) e − z ) ≤ 1 + e − 2 z − 2 e − z + ( 2 − 2 cos ⁡ θ ) e − z =e^{2x}(1+e^{-2z}-2e^{-z}\cos{\theta})=e^{2x}(1+e^{-2z}+(-2+2-2\cos{\theta})e^{-z})\leq 1+e^{-2z}-2e^{-z}+(2-2\cos{\theta})e^{-z}

≤ 1 + e − 2 z − 2 e − z + θ 2 e − z ≤ ( 1 − e − z ) 2 + θ 2 ≤ z 2 + θ 2 = ∣ α − β ∣ 2 . \leq 1+e^{-2z}-2e^{-z}+\theta^2e^{-z} \leq (1-e^{-z})^2+\theta^2 \leq z^2 +\theta^2 = |\alpha-\beta|^2.

James Wilson - 5 months ago

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