Respuesta :
Answer:
Following are the answer to this question:
Step-by-step explanation:
The principle vale of Arg(3)
[tex]Arg(3)=-\pi+\tan^{-1} (\frac{|Y|}{|x|})[/tex]
The principle value of the [tex]\logi= \log(0+i)\ \ \ \ \ _{where} \ \ \ x=0 \ \ y=1> 0[/tex]
So, the principle value:
a)
[tex]\to \log(i)=\log |i|+i Arg(i)\\\\[/tex]
[tex]=\log \sqrt{0+1}+i \tan^{-1}(\frac{1}{0})\\\=\log 1 +i \tan^{-1}(\infty)\\\=0+i\frac{\pi}{2}\\\=i\frac{\pi}{2}[/tex]
b)
[tex]\to \log(-i)= \log(0-i ) \ \ \ x=0 \ \ \ y= -1<0\\[/tex]
Principle value:
[tex]\to \log(-i)= \log|-i|+iArg(-i) \\\\[/tex]
[tex]=\log \sqrt{0+1}+i(-\pi+\tan^{-1}(\infty))\\\\=\log1 + i(-\pi+\frac{\pi}{2})\\\\=-i\frac{\pi}{2}[/tex]
c)
[tex]\to \log(-1+i) \ \ \ \ x=-1, _{and} y=1 \ \ \ x<0 and y>0[/tex]
The principle value:
[tex]\to \log(-1+i)=\log |-1+i| + i Arg(-1+i)[/tex]
[tex]=\log \sqrt{1+1}+i(\pi+\tan^{-1}(\frac{1}{1}))\\\\=\log \sqrt{2} + i(\pi-\tan^{-1}\frac{\pi}{4})\\\\=\log \sqrt{2} + i\tan^{-1}\frac{3\pi}{4}\\\\[/tex]
d)
[tex]\to i^i=w\\\\w=e^{i\log i}[/tex]
The principle value:
[tex]\to \log i=i\frac{\pi}{2}\\\\\to w=e^{i(i \frac{\pi}{2})}\\\\=e^{-\frac{\pi}{2}}[/tex]
e)
[tex]\to (-i)^i\\\to w=(-i)^i\\\\w=e^{i \log (-i)}[/tex]
In this we calculate the principle value from b:
so, the final value is [tex]e^{\frac{\pi}{2}}[/tex]
f)
[tex]\to -1^i\\\\\to w=e^{i log(-1)}\\\\\ principle \ value: \\\\\to \log(-1)= \log |-1|+iArg(-i)[/tex]
[tex]=\log \sqrt{1} + i(\pi-\tan^{-1}\frac{0}{-1})\\\\=\log \sqrt{1} + i(\pi-0)\\\\=\log \sqrt{1} + i\pi\\\\=0+i\pi\\=i\pi[/tex]
and the principle value of w is = [tex]e^{\pi}[/tex]
g)
[tex]\to -1^{-i}\\\\\to w=e^(-i \log (-1))\\\\[/tex]
from the point f the principle value is:
[tex]\to \log(-1)= i\pi\\\to w= e^{-i(i\pi)}\\\\\to w=e^{\pi}[/tex]
h)
[tex]\to \log(-1-i)\\\\\ Here x=-1 ,<0 \ \ y=-1<0\\\\ \ principle \ value \ is:\\\\ \to \log(-1-i)=\log\sqrt{1+1}+i(-\pi+\tan^{-1}(1))[/tex]
[tex]=\log\sqrt{2}+i(-\pi+\frac{\pi}{4})\\\\=\log\sqrt{2}+i(-\frac{3\pi}{4})\\\\=\log\sqrt{2}-i\frac{3\pi}{4})\\[/tex]
