Question

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

Answer 1

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

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


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

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

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


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


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