I am wondering how to compute the line integral of

$$ \mathrm{g}\left(z\right) = z\left(z^{\ast}\right)^{2} - \cos\left(z\right)\quad \mbox{over the curve}\quad \,\mathrm{y}\left(t\right) = \cos\left(2t\right ) + \sin\left(2t\right)\,\mathrm{i}\,,\quad 0 \leq t \leq {\pi \over 2} $$ where $z^{\ast}$ denotes the complex conjugate of $z$.

I am given that the answer is $\pi\mathrm{i} + \sin\left(1\right) - \sin\left(1\right)$

$$ \mbox{What I tried, I tried noting that the integral can be }\quad \int_{0}^{\pi/2}\mathrm{g}\left(\mathrm{y}\left(t\right)\right)\,\mathrm{y}'\left(t\right)\,\mathrm{d}t $$ where $\,\mathrm{g}\left(\mathrm{y}\left(t\right)\right) = \mathrm{e}^{-2\mathrm{i}t} - \cos\left(\mathrm{e}^{2\mathrm{i}t}\right)\quad\mbox{and where}\quad \mathrm{y}'\left(t\right) = -2\sin\left(2t\right) + 2\cos\left(2t\right)\mathrm{i}$.

But then I get stuck on computing next, because I have so many terms multiplying , some containg i and some not, some containing exponentials, setc, so I dont know how to proceed, is there something I am missing? Or can anyone help show me how the answer was gotten? Is this on the right track?


  • $\begingroup$ 1. What curve is parametrised by $y$? 2. Note that $z(z^{\ast})^2 = \dfrac{\lvert z\rvert^4}{z}$. 3. Use the fundamental theorem of calculus for holomorphic functions. $\endgroup$ – Daniel Fischer Sep 27 '16 at 20:54
  • $\begingroup$ The curve is the the top semi circle $\endgroup$ – Quality Sep 27 '16 at 20:57
  • $\begingroup$ Good. Then how does point 2 help here? $\endgroup$ – Daniel Fischer Sep 27 '16 at 20:58
  • $\begingroup$ Since we can write it not contain conjugate we have that it is holomorphic so we can use FTC, but Im having trouble understanding how to integrate that with the modulus $\endgroup$ – Quality Sep 27 '16 at 21:16
  • $\begingroup$ What can you say about $\lvert z\rvert$ on the curve? $\endgroup$ – Daniel Fischer Sep 27 '16 at 21:18

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} z & = \expo{2\ic t}\implies \left\{\begin{array}{rcl} \ds{\mrm{g}\pars{z}} & \ds{=} & \ds{\expo{-2\ic t} - \cos\pars{\expo{2\ic t}}} \\ \ds{\totald{z}{t}} & \ds{=} & \ds{\expo{2\ic t}\pars{2\ic}} \end{array}\right. \end{align} The integral becomes: \begin{align} &\int_{0}^{\pi/2}\bracks{\expo{-2\ic t} - \cos\pars{\expo{2\ic t}}} \bracks{\expo{2\ic t}\pars{2\ic}}\,\dd t = 2\ic\int_{0}^{\pi/2}\,\dd t - \int_{0}^{\pi/2}\cos\pars{\expo{2\ic t}} \bracks{\expo{2\ic t}\pars{2\ic}}\,\dd t \\[5mm] & = 2\ic\,{\pi \over 2} - \bracks{\sin\pars{\expo{2\ic t}}}_{\ 0}^{\ \pi/2} = \pi\ic - \sin\pars{-1} + \sin\pars{1} = \bbox[10px,border:1px groove navy]{\ds{2\sin\pars{1} + \pi\ic}} \end{align}


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