## Exponential complex number question

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

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi Tom! Just write e-2+4i in Cartesian form (ie a + ib), and multiply the whole thing out.
 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

## 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?

Blog Entries: 27
Recognitions:
Gold Member
Homework Help
 Quote by thomas49th $$=> (1-12i)e^{-2}(cos(4^{c})+isin(4^{c}))$$