Find the real part of the complex number: [tex](1-12i)e^{-2+4i}[/tex] I know that z = a + ib can be rewritten as [tex]z = |z|e^{i\theta}[/tex] but that doesn't help because the coefficient of the e is not a scalar value, rather a complex number. Thanks Tom
So, z = a + ib [tex] (1-12i)e^{-2+4i} [/tex] [tex] => (1-12i)e^{-2}e^{4i} [/tex] we can say [tex] e^{4i} = (cos(4^{c})+isin(4^{c})) [/tex] all good so far? so [tex] => (1-12i)e^{-2}(cos(4^{c})+isin(4^{c})) [/tex] we can write the second part of the term in Cartesian (or first part in polar): [tex](1-12i)\cdot e^{-2}\cdot(\frac{1+4}{\sqrt{17}})[/tex] Something is not right as the angle part (b/a) can be ambiguous - 8/2, 4/1, 16/4 ect Is that the correct answer? Thanks Thomas
not sure if this will help: setting up simultaneous equations: [tex]1 = \sqrt(b^{2} + a^{2})[/tex] [tex]1 = b^{2} + a^{2}[/tex] and also [tex]tan\theta = \frac{b}{a}[/tex] so [tex]b = atan\theta[/tex] and [tex]b^{2} = 1 - a^{2}tan\theta[/tex] That any good?
Yes (but what's ^{c} ? … 4 is in radians). Now just multiply it out, and only use the bits of the result that don't have i.