complex number!

**lawochekel** · April 9th 2012, 01:59 AM

find the principal value of log(1+i)

i don't understand the problem above, pls help, thanks.

**princeps** · April 9th 2012, 02:22 AM

$z=\ln(1+i)$

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

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

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

**Prove It** · April 9th 2012, 02:33 AM

Originally Posted by lawochekel

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 $\displaystyle \begin{align*} x + i\,y = \log{(1 + i)} \end{align*}$ , then that's the same as $\displaystyle \begin{align*} e^{x + i\,y} = 1 + i \end{align*}$ .

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

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

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

Substituting into the first equation gives

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

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

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

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

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