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(-1 + i)^7 working?

Guest May 28, 2014

Best Answer 

 #3
avatar+20035 
+8

$$\boxed{(-1 + i)^7 \quad? }$$

z=x+iy  $$\boxed{z=-1+i} \quad \Rightarrow x=-1, y=\textcolor[rgb]{1,0,0}{+}1$$

$$r=\sqrt{x^2+y^2}=\sqrt{(-1)^2+1^2}=\sqrt{2}$$

$$\phi = sign(y)*\cos^{-1}{\left(\frac{x}{r}\right)}
=\textcolor[rgb]{1,0,0}{+}\cos^{-1}(\frac{-1}{\sqrt{2}})
=\cos^{-1}(-\frac{\sqrt{2}}{2})=\frac{3}{4}\pi$$

$$\boxed{z=r*e^{i\phi}}$$

$$\\z=\sqrt2*e^{i*\frac{3}{4}\pi} \\
z^7=\left(\sqrt2*e^{i*\frac{3}{4}\pi}\right)^7 \\
z^7=\left(\sqrt2\right)^7*\left(e^{i*\frac{3}{4}\pi}\right)^7 \\
z^7=\left(\sqrt2\right)^7*e^{i*\frac{7*3}{4}\pi} \\
z^7=\left(\sqrt2\right)^7*e^{i*\frac{21}{4}\pi}
= \left(\sqrt2\right)^7*e^{i\left(4\pi+\pi+\frac{1}{4}\pi\textcolor[rgb]{1,0,0}{-4\pi}}\right)\\$$

$$z^7=\left(\sqrt2\right)^7*e^{i\left(\pi+\frac{1}{4}\pi}\right)\\$$

$$\boxed{e^{i\phi}=\cos{\phi}+i\sin{\phi}}$$

$$\\z^7=\left(\sqrt2\right)^7 \left[ \cos{(\pi+\frac{\pi}{4} )}+i*sin{(\pi+\frac{\pi}{4})} \right]\\
z^7=\left(\sqrt2\right)^7
\left[
\underbrace{\cos{\pi}}_{-1}\cos{\frac{\pi}{4}} - \underbrace{\sin{\pi}}_0\sin{\frac{\pi}{4}}
+i\left( \underbrace{\sin{\pi}}_0\cos{\frac{\pi}{4}}+\underbrace{\cos{\pi}}_{-1}\sin{\frac{\pi}{4}} \right)
\right]\\
z^7=\left(\sqrt2\right)^7 \left( -\cos{\frac{\pi}{4}}-i\sin{\frac{\pi}{4}}\right)$$

$$\sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}=\frac{\sqrt{2} }{2}\quad\\
z^7=\left(\sqrt2\right)^7 \left( -\frac{\sqrt{2}}{2}-i{\frac{\sqrt{2}}{2}\right)\\
z^7=\left(\sqrt2\right)^7\frac{\sqrt{2}}{2}\left( -1-i\right)\\
z^7=\frac{\left(\sqrt{2}\right)^8}{2}\left(-1-i\right)$$

$$\\z^7=\frac{2^4}{2}\left(-1-i\right)\\\\
z^7=2^3\left(-1-i\right)\\
z^7=8\left(-1-i\right)\\
z^7=-8-8i\\$$

$$\boxed{(-1+i)^7=-8-8i}$$

heureka  May 28, 2014
 #1
avatar+93691 
+8

Interesting!

$$^7C_0(-1)^0(i)^7+
^7C_1(-1)^1(i)^6+
^7C_2(-1)^2(i)^5+
^7C_3(-1)^3(i)^4+
^7C_4(-1)^4(i)^3+
^7C_5(-1)^5(i)^2+
^7C_6(-1)^6(i)^1+
^7C_7(-1)^7(i)^0\\\\$$

 

$$=(i)^7+
7(-1)^1(i^2)^3+
^7C_2*1*(i^4)i+
^7C_3*-1*(i^2)^2+
^7C_4*1*(i^2)i+
^7C_5*-1*(i^2)+
7*1(i)^1+-1\\\\$$

 

$$=(-1)i+
7*-1*(-1)^3+
^7C_2*1*(-1)i+
^7C_3*-1*(-1)^2+
^7C_4*1*(-1)i+
^7C_5*-1*(-1)+
7*1*(i)^1+
-1\\\\$$

 

$$=-i+7+^7C_2(i)+
^7C_3(-1)+
^7C_4(-i)+
^7C_5(-1)+
7i
-1\\\\$$

 

$$=-i+7+21i
-35
-i35
+21
+7i
-1\\\\
=i(-1+21-34+7-1)+(7-35+21-1)\\\\
=i(-8)-8\\\\
=-8-8i$$

 

NOTE: I would actually be surprised if this is not riddled with careless errors. (I have fixed one little error)

BUT that is the technique that i would use.

Thanks admin.  

Melody  May 28, 2014
 #2
avatar+3077 
+5

$${\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{i}\right)}^{{\mathtt{7}}} = {\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{8}}{i}$$

$${expand}{\left({\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{i}\right)}^{{\mathtt{7}}}\right)} = {{i}}^{{\mathtt{7}}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{6}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{5}}}{\mathtt{\,-\,}}{\mathtt{35}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{35}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{21}}{\mathtt{\,\times\,}}{{i}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}$$

admin  May 28, 2014
 #3
avatar+20035 
+8
Best Answer

$$\boxed{(-1 + i)^7 \quad? }$$

z=x+iy  $$\boxed{z=-1+i} \quad \Rightarrow x=-1, y=\textcolor[rgb]{1,0,0}{+}1$$

$$r=\sqrt{x^2+y^2}=\sqrt{(-1)^2+1^2}=\sqrt{2}$$

$$\phi = sign(y)*\cos^{-1}{\left(\frac{x}{r}\right)}
=\textcolor[rgb]{1,0,0}{+}\cos^{-1}(\frac{-1}{\sqrt{2}})
=\cos^{-1}(-\frac{\sqrt{2}}{2})=\frac{3}{4}\pi$$

$$\boxed{z=r*e^{i\phi}}$$

$$\\z=\sqrt2*e^{i*\frac{3}{4}\pi} \\
z^7=\left(\sqrt2*e^{i*\frac{3}{4}\pi}\right)^7 \\
z^7=\left(\sqrt2\right)^7*\left(e^{i*\frac{3}{4}\pi}\right)^7 \\
z^7=\left(\sqrt2\right)^7*e^{i*\frac{7*3}{4}\pi} \\
z^7=\left(\sqrt2\right)^7*e^{i*\frac{21}{4}\pi}
= \left(\sqrt2\right)^7*e^{i\left(4\pi+\pi+\frac{1}{4}\pi\textcolor[rgb]{1,0,0}{-4\pi}}\right)\\$$

$$z^7=\left(\sqrt2\right)^7*e^{i\left(\pi+\frac{1}{4}\pi}\right)\\$$

$$\boxed{e^{i\phi}=\cos{\phi}+i\sin{\phi}}$$

$$\\z^7=\left(\sqrt2\right)^7 \left[ \cos{(\pi+\frac{\pi}{4} )}+i*sin{(\pi+\frac{\pi}{4})} \right]\\
z^7=\left(\sqrt2\right)^7
\left[
\underbrace{\cos{\pi}}_{-1}\cos{\frac{\pi}{4}} - \underbrace{\sin{\pi}}_0\sin{\frac{\pi}{4}}
+i\left( \underbrace{\sin{\pi}}_0\cos{\frac{\pi}{4}}+\underbrace{\cos{\pi}}_{-1}\sin{\frac{\pi}{4}} \right)
\right]\\
z^7=\left(\sqrt2\right)^7 \left( -\cos{\frac{\pi}{4}}-i\sin{\frac{\pi}{4}}\right)$$

$$\sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}=\frac{\sqrt{2} }{2}\quad\\
z^7=\left(\sqrt2\right)^7 \left( -\frac{\sqrt{2}}{2}-i{\frac{\sqrt{2}}{2}\right)\\
z^7=\left(\sqrt2\right)^7\frac{\sqrt{2}}{2}\left( -1-i\right)\\
z^7=\frac{\left(\sqrt{2}\right)^8}{2}\left(-1-i\right)$$

$$\\z^7=\frac{2^4}{2}\left(-1-i\right)\\\\
z^7=2^3\left(-1-i\right)\\
z^7=8\left(-1-i\right)\\
z^7=-8-8i\\$$

$$\boxed{(-1+i)^7=-8-8i}$$

heureka  May 28, 2014
 #4
avatar+27061 
+5

$$$$(-1+i)^7=(-1+i)^2(-1+i)^2(-1+i)^2(-1+i)$$\\
$$(-1+i)^2 = 1-2i+i^2=1-2i-1=-2i$$\\
Therefore $$(-1+i)^7=(-2i)^3(-1+i)=(-2)^3i^3(-1+i)=8i(-1+i)=-8-8i$$

Alan  May 28, 2014
 #5
avatar+93691 
0

WOW Alan,

I sure know how to make a mountain out of a mole hill!

I'd double you score it I could!

Melody  May 29, 2014

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