What kind of singularity does \(e^{1/z}\) have at \(z=0\)?

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Review the concept: Isolated singularity

An isolated singularity of a function \(f\) is a point \(z_0\) where \(f\) is not holomorphic but is holomorphic on a punctured disc \(0<|z-z_0|<r\) around it.

More practice problems: isolated singularity

• ZW-ISOLATED-SINGULARITY-001 What kind of singularity does (sin z)/(z) have at z=0?
• ZW-ISOLATED-SINGULARITY-003 What kind of singularity does 1z^2+1 have at z=i?
• ZW-ISOLATED-SINGULARITY-004 The singularity of (cos z-1)/(z^2) at z=0 is removable. What value must f(0) take to remove it?
• ZW-ISOLATED-SINGULARITY-005 What is the order of the pole of 1z^2(z-1) at z=0?
• ZW-ISOLATED-SINGULARITY-006 If |f| is bounded on a punctured disc 0<|z-z_0|<r, what kind of isolated singularity does f have at…
• ZW-ISOLATED-SINGULARITY-007 Which of these functions has at z=0 a singular point that is NOT an isolated singularity?
• ZW-ISOLATED-SINGULARITY-008 Classify the singularity of (z)/(sin z) at z=π.
• ZW-ISOLATED-SINGULARITY-009 Classify the singularity of zsin 1z at z=0.

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