arytmetyka z teoria liczb

[kingula_36]

1.Obliczyc: \((1+i)^{8}\)

2. Wykazac ze \(z^{2}\)=\(w^{2\) wtedy i tylko wtedy z=w lub z=-w

[ewelawwy]

2.
\(z=a+bi\\
w=c+di\\
z^2=a^2+2abi-b^2\\
w^2=c^2+2cdi-d^2\\
z^2=w^2\; \Leftrightarrow \; a^2+2abi-b^2=c^2+2cdi-d^2\; \Leftrightarrow \; a^2-b^2=c^2-d^2\; \wedge \; 2ab=2cd\\
\{a^2-b^2=c^2-d^2\\
2ab=2cd\)
\(a=\frac{cd}b\\
\frac{c^2d^2}{b^2}-b^2=c^2-d^2/\cdot b^2\\
c^2d^2-b^4=c^2b^2-d^2b^2\\
b^4+b^2(c^2-d^2)-c^2d^2=0\\
b^2=t,\; t\ge 0\\
t^2+(c^2-d^2)t-c^2d^2=0\\
\Delta=c^4-2c^2d^2+d^4+4c^2d^2=c^4+d^4+2c^2d^2=(c^2+d^2)^2\\
t_1=\frac{d^2-c^2-c^2-d^2}{2}=-c^2<0\\
t_2=\frac{d^2-c^2+c^2+d^2}{2}=d^2\\
b^2=d^2\\
b=d\; \vee \; b=-d\\
a=\frac{cd}{d}=c \; \vee \; a=\frac{cd}{-d}=-c\\
z=c+di=w \; \vee \; z=-c-di=-(c+di)=-w\)

[irena]

1.
Można tak:
\((1+i)^8=[(1+i)^2]^4=[1+2i+i^2]^4=[1+2i-1]^4=[(2i)^2]^2=[4i^2]^2=[-4]^2=16\)

Można tak:
\(z=1+i\\\sqrt{1^2+1^2}=\sqrt{2}\\z=\sqrt{2}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=\sqrt{2}(cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4}))\\z^8=(\sqrt{2})^8(cos(8\cdot\frac{\pi}{4})+i sin(8\cdot\frac{\pi}{4}))=16(cos2\pi+i sin2\pi)=16(1+0i)=16\)