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Exponential complex number question 
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#1
Mar1410, 01:50 PM

P: 656

Find the real part of the complex number:
[tex](112i)e^{2+4i}[/tex] I know that z = a + ib can be rewritten as [tex]z = ze^{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 


#2
Mar1410, 02:38 PM

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P: 26,148

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


#3
Mar1410, 02:53 PM

P: 656

So,
z = a + ib [tex] (112i)e^{2+4i} [/tex] [tex] => (112i)e^{2}e^{4i} [/tex] we can say [tex] e^{4i} = (cos(4^{c})+isin(4^{c})) [/tex] all good so far? so [tex] => (112i)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](112i)\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 


#4
Mar1410, 02:56 PM

P: 656

Exponential complex number question
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? 


#5
Mar1410, 03:01 PM

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P: 26,148

Now just multiply it out, and only use the bits of the result that don't have i. 


#6
Mar1410, 03:27 PM

P: 656

Genius :)



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