For which values of $z$ the inequality $|e^{z-1}|<2$ holds

Question

I want to find for which values of $z$ the following inquality holds $$|e^{z-1}|<2$$
what I tried to do is:
$$|e^{z-1}|=|e^{x-1+y\mathbb{i}}|<2$$ $$=e^{x-1}\cdot(\cos(y) + \mathbb{i} \sin(y))$$ OR another thing I tried:
$$z-1=\ln(2) \rightarrow z<\ln(2)+1$$ what is right? and how to continue?
thanks.

why is the real part? because I compare coefficients the coefficient of the Im will be zero ? — Ofir Attia, Dec 16 '13 at 14:10
Well, yes: $$|e^{a+ib}|=|e^a|,|e^{ib}|=e^a$$ since, and you by now should be able to prove this, whenever $,b\in\Bbb R\implies ;|e^{bi}|=1;$ ... — DonAntonio, Dec 16 '13 at 14:12

score 2 · Accepted Answer · answered Dec 16 '13 at 14:03

2

Put $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ , then

$$|e^{z-1}|=|e^{x-1+iy}|=e^{x-1}<2\iff x-1<\log 2\iff x<\log 2 +1$$

and that's all!

answered Dec 16 '13 at 14:03

DonAntonio

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For which values ​​of $z$ the inequality $|e^{z-1}|<2$ holds

1 Answers1

For which values of $z$ the inequality $|e^{z-1}|<2$ holds