Zadania z równań i potęg
: 15 lut 2011, o 16:30
Bardzo prosze o sprawdzenie czy dobrze zrobiłem te 2 zadania :
\(\displaystyle{ a)
\left( 1+i\right) ^{20}
\left| z\right|= \sqrt{2}
1+i= \sqrt{2}\left( cos \frac{ \pi }{4}+isin \frac{ \pi }{4} \right)
\left( 1+i\right) ^{20}= \left( \sqrt{2} \right) ^{20}\left( cos \frac{20 \pi }{4}+isin \frac{20 \pi }{4} \right) = 2^{10}\left( cos5 \pi +isin5 \pi \right)= -1024
b)
z^{4}+16=0
\left| z\right|=2
cos \alpha = \frac{ \pi }{4}
sin \alpha = \frac{ \pi }{4}
w_{0}=2 \left( \frac{ \sqrt{2} }{2}+i \frac{ \sqrt{2} }{2} \right) = \sqrt{2}+i \sqrt{2}
w_{1}=2 \left( -\frac{ \sqrt{2} }{2}+i \frac{ \sqrt{2} }{2} \right) = -\sqrt{2}+i \sqrt{2}
w_{2}=2 \left( -\frac{ \sqrt{2} }{2}-i \frac{ \sqrt{2} }{2} \right) = -\sqrt{2}-i \sqrt{2}
w_{3}=2 \left( \frac{ \sqrt{2} }{2}-i \frac{ \sqrt{2} }{2} \right) = \sqrt{2}-i \sqrt{2}}\)
\(\displaystyle{ a)
\left( 1+i\right) ^{20}
\left| z\right|= \sqrt{2}
1+i= \sqrt{2}\left( cos \frac{ \pi }{4}+isin \frac{ \pi }{4} \right)
\left( 1+i\right) ^{20}= \left( \sqrt{2} \right) ^{20}\left( cos \frac{20 \pi }{4}+isin \frac{20 \pi }{4} \right) = 2^{10}\left( cos5 \pi +isin5 \pi \right)= -1024
b)
z^{4}+16=0
\left| z\right|=2
cos \alpha = \frac{ \pi }{4}
sin \alpha = \frac{ \pi }{4}
w_{0}=2 \left( \frac{ \sqrt{2} }{2}+i \frac{ \sqrt{2} }{2} \right) = \sqrt{2}+i \sqrt{2}
w_{1}=2 \left( -\frac{ \sqrt{2} }{2}+i \frac{ \sqrt{2} }{2} \right) = -\sqrt{2}+i \sqrt{2}
w_{2}=2 \left( -\frac{ \sqrt{2} }{2}-i \frac{ \sqrt{2} }{2} \right) = -\sqrt{2}-i \sqrt{2}
w_{3}=2 \left( \frac{ \sqrt{2} }{2}-i \frac{ \sqrt{2} }{2} \right) = \sqrt{2}-i \sqrt{2}}\)