Solved Problem on Contours
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\( \mathsf{c)}\;\; \displaystyle z=\frac{1}{t}+it\qquad ,\qquad 1\leqslant t\lt \infty \)


The function z is a parametric function of the type
\[ \bbox[#99CCFF,10px] {z(t)=x(t)+iy(t)} \]
Identifying the functions x(t) and y(t)
\[ \begin{align} & x(t)=\frac{1}{t} \tag{I}\\[10pt] & y(t)=t \tag{II} \end{align} \]
substituting expression (II) into expression (I)
\[ \begin{gather} x=\frac{1}{y}\\ y=\frac{1}{x} \end{gather} \]
Graph 1

The function z(t) represents a segment of hyperbola oriented from the point (1, 1) to (0, ∞), (Graph 1).

Note: About curve orientation.
When the parameter t is equal to 1, y = t = 1, and \( x=\frac{1}{t}=\frac{1}{1}=1 \), we have the point (1, 1), (Graph 2).

Graph 2

When parameter t is equal to 2, y = t = 2, and   \( x=\frac{1}{t}=\frac{1}{2}=0.5 \), the curve is oriented towards the point (0.5; 2), (Graph 3).

Graph 3

When parameter t is equal to 4, y = t = 4, and   \( x=\frac{1}{t}=\frac{1}{4}=0.25 \), the curve is oriented towards the point (0.25; 4), (Graph 4).

Graph 4

When the parameter t varies from 1 to infinity, the curve is traversed in the orientation that x varies from 1 and tends to 0, and y varies from 1 and tends to infinity.
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