Learn how to simplify any power of the imaginary unit i. For example, simplify i²⁷ as -i.
Log in satyanash 8 years agoPosted 8 years ago. Direct link to satyanash's post “How is `i^3 = -i` ?My un...” How is • (52 votes) andrewp18 8 years agoPosted 8 years ago. Direct link to andrewp18's post “On step 3 you did:√𝑎 • ...” On step 3 you did: (167 votes) Jonathan Huang 7 years agoPosted 7 years ago. Direct link to Jonathan Huang's post “Where do you use this in ...” Where do you use this in the real world? • (15 votes) 🎓 Arnaud Jasperse 7 years agoPosted 7 years ago. Direct link to 🎓 Arnaud Jasperse's post “Physics uses it. Electric...” Physics uses it. Electrical Engineering often have to use the imaginary unit in their calculations, but it is also used in Robotics. There, to rotate an object through 3 dimensions they even result in using Quaternions (which build on the imaginary unit i). (37 votes) dollyrauh 8 years agoPosted 8 years ago. Direct link to dollyrauh's post “Hi, could someone maybe h...” Hi, could someone maybe help me with what I did wrong in the last challenge problem? I followed along with their explanation well enough, but I'm scratching my head as to exactly where I messed up the problem. My answer was i, while the correct was -i. Also, by way of explanation, in steps three, four, and five, I am assuming that 1 can be rewritten as 1=sqrt(1) because 1*1= 1, therefore the square root of 1 equals 1. i^-1= 1/i^1 (because properties of negative exponents) It's driving me crazy that I can't figure out what I did wrong!! Please help. Thanks so much!! • (14 votes) Dominik.Ehlert 8 years agoPosted 8 years ago. Direct link to Dominik.Ehlert's post “It seems to me your probl...” It seems to me your problem is the definition (33 votes) joey mizrahi 8 years agoPosted 8 years ago. Direct link to joey mizrahi's post “how is i^-1 = -idoesnt i...” how is i^-1 = -i • (7 votes) Jacky Xie 8 years agoPosted 8 years ago. Direct link to Jacky Xie's post “Yes, i^-1 = 1/i.Notice t...” Yes, i^-1 = 1/i. (22 votes) Rob 8 years agoPosted 8 years ago. Direct link to Rob's post “will i^0=1?” will i^0=1? • (9 votes) Abhishek Kumar 5 years agoPosted 5 years ago. Direct link to Abhishek Kumar's post “Hey, the explanation to y...” Hey, the explanation to your query is "any number other than 0 taken to the power 0 is defined to be 1". (6 votes) Peu255 a year agoPosted a year ago. Direct link to Peu255's post “where are the powers of t...” where are the powers of the imaginary unit? I've seen no powers. Does it throw fireballs? • (12 votes) O Madhu 💞 a year agoPosted a year ago. Direct link to O Madhu 💞's post “I understand how to solve...” I understand how to solve for i using simple negative exponents, but I don't know how you would solve for more larger negative exponents. • (1 vote) Tanner P a year agoPosted a year ago. Direct link to Tanner P's post “I’ll walk you through an ...” I’ll walk you through an example. Say you want to evaluate i^-207 1. Divide by four and find the remainder Since the remainder is 3, i^-207 = i^-3 To see why this works: Using exponent properties, you can rewrite the problem as 2. Now, this is something you know how to solve i^-207 = i (20 votes) umairansari16 8 years agoPosted 8 years ago. Direct link to umairansari16's post “what Will be i to the pow...” what Will be i to the power i 899999 • (4 votes) kubleeka 8 years agoPosted 8 years ago. Direct link to kubleeka's post “i^899999(i^900000)/i1/i...” i^899999 (12 votes) owengeorge a year agoPosted a year ago. Direct link to owengeorge's post “Is there any reason to us...” Is there any reason to use multiple of 4 over multiple of 1,2 etc or is just arbitrary? • (4 votes) Kim Seidel a year agoPosted a year ago. Direct link to Kim Seidel's post “It must be 4. It is not a...” It must be 4. It is not arbitrary. The powers of i rotate thru 4 values, not something else. Hope this helps. (11 votes) tue99452 7 years agoPosted 7 years ago. Direct link to tue99452's post “what is the answer of i ^...” what is the answer of i ^ i ? • (6 votes) Ron Joniak 7 years agoPosted 7 years ago. Direct link to Ron Joniak's post “Well, due to a property t...” Well, due to a property that you might learn later, i = e^(i*pi/2). So, i^i = e^(i*pi/2*i). We know i*i = -1, so i^i = e^(-pi/2). i^i is always a real number, but there are some problems with this definition based on the property. (4 votes)Want to join the conversation?
i^3 = -i ?
My understanding leads me to:i^3 = i * i * i
i^3 = sqrt( -1 ) * sqrt( -1 ) * sqrt( -1 ) ; i = sqrt( -1 )
i^3 = sqrt( -1 * -1 * -1 )
i^3 = sqrt( -1 )
i^3 = i
√𝑎 • √𝑏 = √(𝑎𝑏)
But that is only true if:
𝑎, 𝑏 > 0
Which is not the case here. Comment if you want to see a proof of that.
1/i (because i^1=i)
1/sqrt(-1) (because i=sqrt(-1))
sqrt(1)/sqrt(-1) (because the 1 in the numerator can be rewritten as sqrt(1) without changing its value)
sqrt(1/-1) (to simplify inside the square root)
sqrt(-1) (because 1/-1 is -1)
sqrt(-1)=i (because i=sqrt(-1))
=ii = sqrt(-1). Better use i^2 = -1. You can see, that your last step would get you sqrt(-1) = +/- i, so the right solution is at least within, but the problem is better solved another way:i ^ (-1) = 1/i = 1/i * i/i <-- Expand with i = i/(i^2) <-- Multiply= i/(-1) <-- Definition= -i <-- Simplify
Therefore i^(-1) = -i
doesnt it equal 1/i
Notice that you can multiply 1/i by i/i, which gives you i/-1, or -i, which is much easier to comprehend than 1/i.
So i=√-1, "i" here is a real number. So i^0=1.
It is explained in great detail at this website-->
(though it requires some math, though eventually, you get the idea)
http://mathworld.wolfram.com/Power.html
However, "zero to the power of zero" (or 0^0) is undefined.
207/4 = 51 R 3
(i^4)^-51 * i^-3
=1^-51 * i^-3
=1 * i^-3
=i^-3
i^-3
=1/i^3
=1/-i
=1/-i * i/i
=i/-i^2
=i/-(-1)
=i/1
=i.
(i^900000)/i
1/i
i/(i^2)
i/(-1)
-i
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
Then this pattern repeats
i^5 = i
i^6 = -1
i^7 = -i
i^8 = 1
etc.
