Exercise · ZW-CURVE-AND-CONTOUR-008 · ★★★ Advanced
Is \(\gamma(t)=t^3+it^3\) a smooth curve on \([-1,1]\)? Its image is a segment.
Choose one option.
A Yes: it is a polynomial and all polynomials are smooth in this sense B No: \(\gamma'(0)=0\), and the definition requires a nonzero derivative C No: the image is a segment, not a curve D Yes: its derivative \(3t^2(1+i)\) is continuous
Show answer B. No: \(\gamma'(0)=0\), and the definition requires a nonzero derivative
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Explanation A continuous \(\gamma'\) is not enough: \(\gamma'(t)=3t^2(1+i)\) vanishes at \(t=0\). The image is a segment, but this parametrization is not smooth; \(\gamma(t)=t+it\) traces the same segment and is smooth.
A smooth curve \(\gamma\) is a function from a real interval \([a,b]\) into \(\mathbb{C}\) with continuous, nonzero derivative ; a contour is a finite chain of smooth curves joined end to end.
\[\gamma(t)=x(t)+i\,y(t),\qquad a\le t\le b\]
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