Комплексни числа

[Гост]

Представете в тригонометричен вид числата:
а) −1 − i
б) [tex]\frac{1-i \sqrt{3} }{2}[/tex]
в) [tex]\frac{2}{1-i \sqrt{3} }[/tex]
г) [tex]\frac{1-i}{1+i}[/tex]

[ammornil]

Представете в тригонометричен вид числата:
а) −1 − i
б) [tex]\frac{1-i \sqrt{3} }{2}[/tex]
в) [tex]\frac{2}{1-i \sqrt{3} }[/tex]
г) [tex]\frac{1-i}{1+i}[/tex]

Знаем, че [tex]\cos(- \alpha )=\cos{ \alpha }, \phantom{QQ} \sin(- \alpha )=-\sin{ \alpha }[/tex]

a)
[tex]A=-1-i \rightarrow III кв.[/tex]
[tex]A=-1-i, \phantom{QQ} |A|=\sqrt{(-1)^{2}+(-1)^{2}}=\sqrt{2}, \phantom{QQ} \tg(\varphi _{A})=\frac{-1}{-1}=1 \Rightarrow \begin{cases}\varphi _{A}=\frac{\pi}{4}\ [rad]\\ \underline{ \varphi _{A}=-\frac{3\pi}{4}\ [rad]} \rightarrow III кв. \end{cases}[/tex]
[tex]A=|A|[\cos(\varphi _{A})+i\sin(\varphi _{A})]=\sqrt{2}\left[\cos{\left(-\frac{3\pi}{4}\right)}+i\sin{\left(-\frac{3\pi}{4}\right)} \right][/tex]

б)
[tex]A= \frac{1}{2}-\frac{\sqrt{3}}{2}i \rightarrow IVкв.[/tex]
[tex]A=\frac{1}{2}-\frac{\sqrt{3}}{2}i, \phantom{QQ} |A|=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(-\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{\frac{4}{4}}=1, \phantom{QQ} \tg(\varphi _{A})=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3} \Rightarrow \begin{cases}\varphi _{A}=\frac{2\pi}{3}\ [rad]\\ \underline{ \varphi _{A}=-\frac{\pi}{3}\ [rad]} \rightarrow IV кв. \end{cases}[/tex]
[tex]A=|A|[\cos(\varphi _{A})+i\sin(\varphi _{A})]=1\left[\cos{\left(-\frac{\pi}{3}\right)}+i\sin{\left(-\frac{\pi}{3}\right)} \right][/tex]

в)
[tex]A= \frac{2}{1-i\sqrt{3}}.\frac{1+i\sqrt{3}}{1+i\sqrt{3}} = \frac{2+2i\sqrt{3}}{1^{2}-(i\sqrt{3})^{2}}=\frac{2+2i\sqrt{3}}{1+3}=\frac{2+2i\sqrt{3}}{4} = \frac{1}{2}+\frac{\sqrt{3}}{2}i \rightarrow Iкв.[/tex]
[tex]A=\frac{1}{2}+\frac{\sqrt{3}}{2}i, \phantom{QQ} |A|=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{\frac{4}{4}}=1, \phantom{QQ} \tg(\varphi _{A})=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \Rightarrow \begin{cases}\varphi _{A}= \underline{ \frac{\pi}{3}\ [rad] } \rightarrow I кв.\\ \varphi _{A}=-\frac{2\pi}{3}\ [rad] \end{cases}[/tex]
[tex]A=|A|[\cos(\varphi _{A})+i\sin(\varphi _{A})]=1\left[\cos{\left(\frac{\pi}{3}\right)}+i\sin{\left(\frac{\pi}{3}\right)} \right][/tex]

г)
[tex]\frac{1-i}{1+i}.\frac{1-i}{1-i}=\frac{(1-i)^{2}}{1^{2}-(i)^{2}}=\frac{1-2i+i^{2}}{1+1}=\frac{1-2i-1}{2}=0-i \rightarrow[/tex] точка, лежаща на отрицателния клон на ординатата [tex]\Rightarrow \varphi _{A}=-\frac{\pi}{2}, \phantom{QQ} \nexists \tg(\varphi _{A})[/tex]
[tex]|A|=\sqrt{0^{2}+(-1)^{2}}=1[/tex]
[tex]A=|A|[\cos(\varphi _{A})+i\sin(\varphi _{A})]=1\left[\cos{\left(-\frac{\pi}{2}\right)}+i\sin{\left(-\frac{\pi}{2}\right)} \right][/tex]