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Thread: complex number!

  1. #1
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    complex number!

    find the principal value of log(1+i)

    i don't understand the problem above, pls help, thanks.
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  2. #2
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    Re: complex number!

    $\displaystyle z=\ln(1+i)$

    $\displaystyle z=\ln\left(\sqrt 2 \cdot e^{i \cdot \frac{\pi}{4}}\right)$

    $\displaystyle z=\ln \sqrt 2 +i\cdot\frac{ \pi}{4}$

    $\displaystyle z=\frac{\ln 2}{2}+i\cdot \frac{\pi}{4}$
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  3. #3
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    Re: complex number!

    Quote Originally Posted by lawochekel View Post
    find the principal value of log(1+i)

    i don't understand the problem above, pls help, thanks.
    Well, it's asking you to write this function in terms of its real and imaginary parts. Anyway, if $\displaystyle \displaystyle \begin{align*} x + i\,y = \log{(1 + i)} \end{align*}$, then that's the same as $\displaystyle \displaystyle \begin{align*} e^{x + i\,y} = 1 + i \end{align*}$.

    $\displaystyle \displaystyle \begin{align*} e^{x + i\,y} &= 1 + i \\ e^xe^{i\,y} &= 1 + i \\ e^x\left[\cos{(y)} + i\sin{(y)}\right] &= 1 + i \\ e^x\cos{(y)} + i\,e^x\sin{(y)} &= 1 + i \end{align*}$

    Equating real and imaginary parts gives $\displaystyle \displaystyle \begin{align*} e^x\cos{(y)} = 1 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} e^x\sin{(y)} = 1 \end{align*}$. Solving these simultaneously by dividing the first equation by the second gives

    $\displaystyle \displaystyle \begin{align*} \frac{e^x\sin{(y)}}{e^x\cos{(y)}} &= \frac{1}{1} \\ \tan{(y)} &= 1 \\ y &= \arctan{(1)} + \pi n, n \in \mathbf{Z} \\ y &= \frac{\pi}{4} + \pi n \end{align*}$

    Substituting into the first equation gives

    $\displaystyle \displaystyle \begin{align*} e^x\cos{(y)} &= 1 \\ e^x\cos{\left(\frac{\pi}{4} + \pi n\right)} &= 1 \\ \frac{(-1)^n e^x}{\sqrt{2}} &= 1 \\ e^x &= \frac{\sqrt{2}}{(-1)^n} \\ x &= \ln{\left[\frac{\sqrt{2}}{(-1)^n}\right]} \end{align*}$

    But since $\displaystyle \displaystyle \begin{align*} x \end{align*}$ is real, that means everything inside the logarithm can only be positive, therefore we can only accept EVEN values of $\displaystyle \displaystyle \begin{align*} n \end{align*}$, so we'll write $\displaystyle \displaystyle \begin{align*} n = 2m, m \in \mathbf{Z} \end{align*}$, which gives us $\displaystyle \displaystyle \begin{align*} x = \ln{\left(\sqrt{2}\right)} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} y = \frac{\pi}{4} + 2\pi m \end{align*}$.

    Therefore $\displaystyle \displaystyle \begin{align*} \log{(1 + i)} = \ln{\left(\sqrt{2}\right)} + i\left(\frac{\pi}{4} + 2\pi m\right) \end{align*}$, and since this is multi-valued, we define the principal value as simply being the "first" possible value (in this case, where $\displaystyle \displaystyle \begin{align*} m = 0 \end{align*}$).

    So the principal value of $\displaystyle \displaystyle \begin{align*} \log{(1 + i)} = \ln{\left(\sqrt{2}\right)} + i\,\frac{\pi}{4} \end{align*}$.
    Thanks from lawochekel
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