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Thread: Complex numbers

  1. #1
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    Complex numbers

    This is for a Principles of Applied Mathematic college course.

    I am having trouble expressing ln(-1-i) in the form a+ib

    Thank you
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  2. #2
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    Re: Complex numbers

    What do you mean by "having trouble"? Have you tried anything at all on this? What do you understand about this? There is a general formula for ln(x+ iy) but I presume you do not know that. Do you known how to write a+ bi in "polar" or "exponential form, $\displaystyle re^{i\theta}$? If you write $\displaystyle a+ib$ in the form $\displaystyle re^{i\theta}$, then $\displaystyle ln(a+ bi)= ln(re^{i\theta})= ln(r)+ i\theta$.
    (That is the "principle value". Since $\displaystyle \theta+ 2k\pi$ is the same as angle $\displaystyle \theta$ for any integer, k, the general formula is $\displaystyle ln(re^{i\theta})= ln(r)+ i(\theta+ 2k\pi)$ for any integer, k.)
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  3. #3
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    Re: Complex numbers

    Ok. yes I do know polar, i was just unclear on what direction to take this. So now Ive gotten:

    ln(sqrt(2))-i135
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    Re: Complex numbers

    You can't use degrees for complex numbers. You have to use radians.
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    Re: Complex numbers

    so then ln(sqrt(2))-i(3pi/4)
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    Re: Complex numbers

    Quote Originally Posted by Tlmetcalfe View Post
    so then ln(sqrt(2))-i(3pi/4)
    that's the value closest to the origin, yes.

    $\ln(z)$ is multivalued

    you might want to fully simplify it to

    $\dfrac 1 2 \ln(2) - i \dfrac {3\pi}{4}$

    and then including the multiple values we can write

    $\ln(-1-i) = \dfrac 1 2 \ln(2) - i\left(\dfrac{3\pi}{4} \pm 2\pi k\right),~~k \in \mathbb{Z}$
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  7. #7
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    Re: Complex numbers

    Quote Originally Posted by Tlmetcalfe View Post
    This is for a Principles of Applied Mathematic college course.

    I am having trouble expressing ln(-1-i) in the form a+ib

    Thank you
    $\displaystyle \begin{align*} z &= \ln{ \left( -1 - \mathrm{i} \right) } \\ \mathrm{e}^{z} &= -1 - \mathrm{i} \\ \mathrm{e}^z &= \sqrt{2}\,\mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i} } \textrm{ where } n \in \mathbf{Z} \\ \mathrm{e}^{ x + \mathrm{i}\,y } &= \sqrt{2}\,\mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\, n \right) \,\mathrm{i} } \\ \mathrm{e}^x\,\mathrm{e}^{ \mathrm{i}\,y } &= \sqrt{2}\,\mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i} } \end{align*}$

    so $\displaystyle \begin{align*} \mathrm{e}^{x} = \sqrt{2} \implies x = \ln{ \left( \sqrt{2} \right) } = \frac{1}{2}\ln{ \left( 2 \right) } \end{align*}$ and $\displaystyle \begin{align*} \mathrm{e}^{ \mathrm{i}\,y } = \mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i} } \implies y = -\frac{3\pi}{4} + 2\,\pi\,n \end{align*}$

    Thus $\displaystyle \begin{align*} z = \frac{1}{2}\ln{ \left( 2 \right) } + \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \, \mathrm{i} \end{align*}$.
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  8. #8
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    Re: Complex numbers

    Quote Originally Posted by Tlmetcalfe View Post
    Ok. yes I do know polar, i was just unclear on what direction to take this. So now Ive gotten:
    ln(sqrt(2))-i135
    Is much as I hate it, yes that is what most applied mathematics textbooks teach as the correct answer to your question.
    Only from my experience, they would expect $\displaystyle 0.5\ln(2)-135\bf{i}$ as the correct form.

    @others, note that the question requires the $a+b\bf{i}$ form.
    That implies that $a~\&~b$ are singular real numbers.
    Last edited by Plato; Mar 28th 2016 at 02:49 PM.
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  9. #9
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    Re: Complex numbers

    Quote Originally Posted by Plato View Post
    Is much as I hate it, yes that is what most applied mathematics textbooks teach as the correct answer to your question.
    Only from my experience, they would expect $\displaystyle 0.5\ln(2)-135\bf{i}$ as the correct form.
    Really? Not $\displaystyle 0.5\ln(2)- (3\pi/4)\bf{i}$? I have never seen a text book, applied mathematics or otherwise, use degrees rather than radians here.

    @others, note that the question requires the $a+b\bf{i}$ form.
    That implies that $a~\&~b$ are singular real numbers.
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  10. #10
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    Re: Complex numbers

    Quote Originally Posted by HallsofIvy View Post
    Really? Not $\displaystyle 0.5\ln(2)- (3\pi/4)\bf{i}$? I have never seen a text book, applied mathematics or otherwise, use degrees rather than radians here.
    The only time I've ever seen degrees used for something similar to this is with phasor notation.

    It wasn't totally uncommon to see for example

    $1+i = \sqrt{2} \angle 45$

    and this was in EE textbooks, not Math.
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  11. #11
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    Re: Complex numbers

    Yes, but then those darned engineers use "j" rather than "i" so what do they know!
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  12. #12
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    Re: Complex numbers

    Quote Originally Posted by HallsofIvy View Post
    Really? I have never seen a text book, applied mathematics or otherwise, use degrees rather than radians here.
    I truly hope that is now the case. I have been out of the fight for ten years. But I hated be threaded with the lost of a large credit hour producing course they want engineers to use degrees.

    Quote Originally Posted by HallsofIvy View Post
    Yes, but then those darned engineers use "j" rather than "i" so what do they know!
    Now that on I never gave in on that. Luckily it only seems to matter a small subset.
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