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avatar+40 

1/i=(1/-1)^0.5=-1^0.5=i

    =i/-1=-i

yomyhomies  Feb 20, 2017

Best Answer 

 #2
avatar+91239 
+10

Thanks Max,

I will try to explain :)

 

 

\(i^0=1\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\~\\ \text{for integer values of n the pattern can be continued}\\ i^{4n}=1\\ i^{4n+1}=i\\ i^{4n+2}=-1\\ i^{4n+3}=-i\\~\\ \text{for n=-1}\\ i^{(4*-1)+3}=i^{-1}=\frac{1}{i}=-i\)

 

Now I will try to explain your logic error when you found i^-1=i

 

When you square a negative number you get a positive number.  When you square root a positive number there are really 2 answers.  By convention we take the positive answer.

See if you can understand this with my example :)

 

\( \frac{1}{-4}=\sqrt{\frac{1}{(-4)^2}}=\frac{1}{4}\qquad \text{obviously not true} \\\frac{1}{i}=\sqrt{\frac{1}{(i)^2}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=i \qquad \text{Also not true}\)

Melody  Feb 20, 2017
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2+0 Answers

 #1
avatar+6900 
+5

\(\boxed{i^{4n-1} = -i}\)

\(1/i\\ =i^{4\cdot 0 - 1}\\ =-i\)

MaxWong  Feb 20, 2017
 #2
avatar+91239 
+10
Best Answer

Thanks Max,

I will try to explain :)

 

 

\(i^0=1\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\~\\ \text{for integer values of n the pattern can be continued}\\ i^{4n}=1\\ i^{4n+1}=i\\ i^{4n+2}=-1\\ i^{4n+3}=-i\\~\\ \text{for n=-1}\\ i^{(4*-1)+3}=i^{-1}=\frac{1}{i}=-i\)

 

Now I will try to explain your logic error when you found i^-1=i

 

When you square a negative number you get a positive number.  When you square root a positive number there are really 2 answers.  By convention we take the positive answer.

See if you can understand this with my example :)

 

\( \frac{1}{-4}=\sqrt{\frac{1}{(-4)^2}}=\frac{1}{4}\qquad \text{obviously not true} \\\frac{1}{i}=\sqrt{\frac{1}{(i)^2}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=i \qquad \text{Also not true}\)

Melody  Feb 20, 2017

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