Solved Problem on Contours
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\( \mathsf{g)}\;\; \displaystyle z=\sqrt{1-t^{2}\;}+it\qquad ,\qquad -1\leqslant t\leqslant 1 \)
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)
\[
\begin{align}
& x(t)=\sqrt{1-t^{2}\;} \tag{I}\\[10pt]
& y(t)=t \tag{II}
\end{align}
\]
substituting expression (I) into expression (II)
\[
x=\sqrt{1-y^{2}\;}
\]
squaring both sides of equation
\[
\begin{gather}
x^{2}=\left(\sqrt{1-y^{2}\;}\right)^{2}\\
x^{2}=1-y^{2}\\
x^{2}+x^{2}=1
\end{gather}
\]
For t = −1, we have x = −1 and
\( x=\sqrt{1-(-1)^{2}\;}=0 \),
for t = 0, we have x = 0 and
\( x=\sqrt{1-0^{2}\;}=1 \),
for t = 1, we have x = 1 and
\( x=\sqrt{1-1^{2}\;}=0 \).
The function z(t) represents a semicircle on the right side of the y-axis (1st and 4th quadrants) and oriented from −1 to 1, (Graph 1).
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