# Exponential complex number question

by thomas49th
Tags: complex, exponential, number
 P: 656 Find the real part of the complex number: $$(1-12i)e^{-2+4i}$$ I know that z = a + ib can be rewritten as $$z = |z|e^{i\theta}$$ but that doesn't help because the coefficient of the e is not a scalar value, rather a complex number. Thanks Tom
 Sci Advisor HW Helper Thanks P: 26,148 Hi Tom! Just write e-2+4i in Cartesian form (ie a + ib), and multiply the whole thing out.
 P: 656 So, z = a + ib $$(1-12i)e^{-2+4i}$$ $$=> (1-12i)e^{-2}e^{4i}$$ we can say $$e^{4i} = (cos(4^{c})+isin(4^{c}))$$ all good so far? so $$=> (1-12i)e^{-2}(cos(4^{c})+isin(4^{c}))$$ we can write the second part of the term in Cartesian (or first part in polar): $$(1-12i)\cdot e^{-2}\cdot(\frac{1+4}{\sqrt{17}})$$ 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
 P: 656 Exponential complex number question not sure if this will help: setting up simultaneous equations: $$1 = \sqrt(b^{2} + a^{2})$$ $$1 = b^{2} + a^{2}$$ and also $$tan\theta = \frac{b}{a}$$ so $$b = atan\theta$$ and $$b^{2} = 1 - a^{2}tan\theta$$ That any good?
 Quote by thomas49th $$=> (1-12i)e^{-2}(cos(4^{c})+isin(4^{c}))$$