Exponential complex number question

thomas49th

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

tiny-tim

Hi Tom!

Just write e^-2+4i in Cartesian form (ie a + ib), and multiply the whole thing out.

thomas49th

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

thomas49th

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?

tiny-tim

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.

thomas49th

Genius :)