+0

# 1/i= -i or i?

+1
125
2
+36

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

=i/-1=-i

yomyhomies  Feb 20, 2017

#2
+90581
+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
Sort:

#1
+6810
+5

$$\boxed{i^{4n-1} = -i}$$

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

MaxWong  Feb 20, 2017
#2
+90581
+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

### 18 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details