Question

Variáveis complexas.

Answer 1

Resposta:

Questão 1 - a)

$\int\limits_{C} {\frac{e^{\pi{z}} }{(z-4i)(z^{2}+1)} } \, dz$

$2\pi{i}[ \lim_{z \to \++i}{\frac{e^{\pi z} }{(z-4i)(z-1)(z+i)}(z-1) } + \lim_{z \to \--i}{\frac{e^{\pi z} }{(z-4i)(z-1)(z+i)}(z+1) } ]\\\\2\pi{i}[ \lim_{z \to \++i}{\frac{e^{\pi z} }{(z-4i)(z-1)(z+i)}(z-1) } + \lim_{z \to \--i}{\frac{e^{\pi z} }{(z-4i)(z-1)(z+i)}(z+1) } ]\\\\2\pi{i}[(\frac{cos\pi +isen\pi }{-3i+2i} )+(\frac{cos\pi -isen\pi}{-5i-2i} )]\\\\2\pi{i}[\frac{1}{i} +\frac{1}{7i} ]\\\\2\pi{i}[\frac{7+1}{7i}]\\\\\frac{16\pi }{7}$

Questão 1 - b)

$\int\limits_{C} {\frac{1}{z^2-1} } \, dz\\\\\int\limits_{C} {\frac{1}{(z-1)(z+1) } \, dz\\\\\int\limits_{C} {\frac{1/2}{z-1} -\frac{1/2}{z+1} } \, dz\\\\\frac{1}{2} \int\limits_{C} {\frac{dz}{z-1} } - \frac{1}{2} \int\limits_{C} {\frac{dz}{z+1} }\\\\\pi i-\pi i=0$

Questão 1

$z=2+i+i(1-2i)+(2+3i)(2+i)+\frac{1-2i}{3+4i}\\\\z=2+i+i-2i^{2}+4+2i+6i+3i^{2}+\frac{1-2i}{3+4i}\\\\z=2+i+i+2+4+2i+6i-3+\frac{1-2i}{3+4i}\\\\z=5+10i+\frac{1-2i}{3+4i}\\\\z=5+10i+(\frac{1-2i}{3+4i}*\frac{3-4i}{3-4i})\\\\z=5+10i+(\frac{3-4i-6i+8i^{2}}{9-12i+12i+16i^{2}} )\\\\z=5+10i+(\frac{3-10i-8}{9+16} )\\\\z=5-10i+\frac{-5-10i}{25} \\\\z=5-10i+\frac{-1-2i}{5}\\\\z=\frac{25+50i-1-2i}{5} \\\\z=\frac{24}{5} +\frac{48i}{5}$

$p^{2}=a^{2}+b^{2}\\\\a=\frac{24}{5}\\\\b=\frac{48}{5} \\\\p^{2}=\frac{576}{25} +\frac{2.304}{25}\\\\p^{2}=\frac{2.880}{25}\\\\p=\sqrt{\frac{2.880}{25}} =\frac{1}{5} \sqrt{2.880} =\frac{1}{5} \sqrt{2^{6}*3^{2}*5}=\frac{2^{3}*3\sqrt{5}}{5}\\\\p=\frac{24\sqrt{5} }{5} \\\\tg\theta=\frac{48/5}{24/5}=\frac{48}{5}*\frac{5}{24}=\frac{48}{24}=2\\\\\theta=arctg(2)=63,43\º$

Convertendo para radianos

$180\º----\pi \\63,43\º---\pi \\\\\\180x=63,43\pi \\\\x=\frac{63,43\pi }{180}\\\\x=0,35\pi$

Assim:

$z=p*e^{i\theta}=>z=\frac{24\sqrt{5} }{5}*e^{0,35\pi{i}}$