- #1

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[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

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- Thread starter thomas49th
- Start date

- #1

- 655

- 0

[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

- #2

tiny-tim

Science Advisor

Homework Helper

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Hi Tom!

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

Just write e

- #3

- 655

- 0

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

- #4

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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

tiny-tim

Science Advisor

Homework Helper

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[tex]

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

[/tex]

Yes (but what's

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

- #6

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Genius :)

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