Представив числа z1=1+i и z2=1-i√3 в тригонометрической форме,вычислить: а) z1*z2 б) z1/z2 в) z^6 г) ^4√z

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Ответ:

valdan2008

02.07.2020 17:25

$z = a + bi, \ z = |z|(\cos(\arg(z)) + i\sin(\arg(z)))\\\\ \ |z| = \sqrt{a^2 + b^2}, \ \arg(z) = arctg(\frac{b}{a}) \\\\ z_1 = 1 + i, \ z_2 = 1 - i\sqrt{3}$

$|z_1| = \sqrt{1^2 + 1^2} = \sqrt{2}, \ \arg z_1 = arctg\frac{1}{1} = arctg1 = \frac{\pi}{4}\\\\ z_1 = \sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})\\\\ |z_2| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{4} = 2, \ \arg z_2 = arctg( -\frac{\sqrt{3}}{1}) = -\frac{\pi}{3}\\\\ z_2 = 2\left(\cos( -\frac{\pi}{3}) + i\sin( -\frac{\pi}{3})\right)$

$a) \ z_1*z_2 = \sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})*2\left(\cos( -\frac{\pi}{3}) + i\sin( -\frac{\pi}{3})\right) = \\\\ = 2\sqrt{2}(\cos \frac{\pi}{4}\cos( -\frac{\pi}{3}) + i\cos \frac{\pi}{4}\sin( -\frac{\pi}{3}) +\\\\ + i\sin \frac{\pi}{4}\cos( -\frac{\pi}{3}) + i^2 \sin \frac{\pi}{4}\sin( -\frac{\pi}{3})) =$

$
 = 2\sqrt{2}(\cos \frac{\pi}{4}\cos( -\frac{\pi}{3}) - \sin \frac{\pi}{4}\sin( -\frac{\pi}{3}) + \\\\ + i(\cos \frac{\pi}{4}\sin( -\frac{\pi}{3}) + \sin \frac{\pi}{4}\cos( -\frac{\pi}{3}))) = \\\\
= 2\sqrt{2}(\cos(\frac{\pi}{4} - \frac{\pi}{3}) + i\sin(\frac{\pi}{4} - \frac{\pi}{3})) =\\\\
= \boxed{2\sqrt{2}\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right)}$

$b) \ \frac{z_1}{z_2} = \frac{ \sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})}{2\left(\cos( -\frac{\pi}{3}) + i\sin( -\frac{\pi}{3})\right)} = \frac{\sqrt{2}}{2}\left(\cos(\frac{\pi}{4} + \frac{\pi}{3}) + i\sin(\frac{\pi}{4} + \frac{\pi}{3}) \right)= \\\\ =\boxed{ \frac{\sqrt{2}}{2}\left(\cos\left(\frac{7\pi}{12}\right) + i\sin\left(\frac{7\pi}{12}\right)\right)}$

$c) \ z_1^6 = (\sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}))^6 = 8(\cos \frac{6\pi}{4} + i\sin \frac{6\pi}{4}) =\\\\ = \boxed{8\left(\cos \left(\frac{3\pi}{2}\right) + i\sin \left(\frac{3\pi}{2}\right)\right)} \\\\\\
z_2^6 = (2\left(\cos( -\frac{\pi}{3}) + i\sin( -\frac{\pi}{3})\right))^6 = 64(\cos (-\frac{6\pi}{3}) + i\sin (-\frac{6\pi}{3})) =\\\\ = \boxed{64\left(\cos (-2\pi) + i\sin (-2\pi)\right)}$

$d) \ \sqrt[4]{ z_1} = \sqrt[4]{\sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})} =\\\\= \boxed{ \sqrt[8]{2}\left(\cos\left( \frac{\pi}{16} + \frac{2k \pi}{4}\right) + i\sin\left( \frac{\pi}{16} + \frac{2k \pi}{4}\right)\right), \ k = 0, 1, 2, 3}\\\\\\
\sqrt[4]{ z_2} = \sqrt[4]{2\left(\cos( -\frac{\pi}{3}) + i\sin( -\frac{\pi}{3})\right)} =\\\\ = \boxed{ \sqrt[4]{2}\left(\cos\left(-\frac{\pi}{12} + \frac{2k \pi}{4}\right) + i\sin\left(-\frac{\pi}{12} + \frac{2k \pi}{4}\right)\right), \ k = 0, 1, 2, 3}
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