Exponential complex number question

[thomas49th]

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

 

[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

$$(1-12i)e^{-2+4i}$$

$$=> (1-12i)e^{-2}e^{4i}$$

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

 

[thomas49th]

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?

 

[tiny-tim]

$$=> (1-12i)e^{-2}(cos(4^{c})+isin(4^{c}))$$

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 :)