what is i^335 i got i?
337374
i^2 = -1\nSo, i^4 = 1.\nSo, i^4n = 1.\n\nSo, i^335 = i^(4*83+3) = i^(4*83)*i^3 = 1*(-1)*i = -i
758984
i^(-1) = 1/i = -i\n\nsummation from i = 1 to n i^(-1) = -ni\n\nput if you mistyped and meant which is likely\n\ni^(-n) it is geometic series\n\n= -(i+i^2+i^3+..... )\n= -(1-i^(n+1))/(1-i)
377581
I am going to assume that you are talking about the following:\n n\n ∑ f(i)\ni=k\nThis means the sum of f(i) for i=k to n.  In other words,  that expression equals f(k)+f(k+1)+f(k+2)+...+f(n-1)+f(n).\n\nThe "pi notation" that I think you must be talking about is this:\n\n n\n Π f(i)\ni=k\nThis means the product of all of the f(i)'s from i=k to n.  In other words, this expression equals:\nf(k)*f(k+1)*...*f(n).\nClearly very similar to the sigma notation.
786717
its 1\n\nsince i = (-1)^1/2\n\nso i^2 = (-1)^(2*(1/2))\n\nwhich is 1
453640
i\nii\niii\niv\nv\nvi
983872
the answer I got was 844093.44N\n\nhere is how I did it\nhttp://img.photobucket.com/albums/v628/kerahna/physyahoo2.jpg\n\nI hope its right XD
440770
i think u mean (7-i)/(1+i)\n\nto solve this u multiply the top and bottom by the complex conjugate of the denominator, so in this case by (1-i)\n\nso we get (7-i)/(1+i) * (1-i)/(1-i) = (7-8i+i^2)/(1-i^2)\n\nwe know i^2 = -1\n\nso answer is (6-8i)/2 = 3-4i
104563
My answer is:\n\n.217N
958486
Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1
864317
I = 1\nV = 5\nX = 10\nC = 100\nD= 500\nM = 1000\n\nMMMCDXIV\n\nSo break it down\n\nMMM = 3*1000\n\nIf a small number precedes a larger one then u take the smaller from the larger\n\nCD = 500-100 = 400\n\nX= 10\n\nIV = 4\n\nSo\n\n3414
954343
Use FOIL and multiply this out just like you would any other algebraic problem, and recall that i² = -1 when solving:\n\n(4-i)(5+i) = \n\n4*5 + 4i - 5i - i² = \n\n20 + 4i - 5i - (-1) =\n\n21 - i\n\nANSWER:  C. 21-i
799993
Use the following to help solve this problem:\n\ni ^1 = i\ni ^2 = -1\ni ^3 = -i\ni ^4 = 1\n\nThis pattern then repeats:\n\ni ^5 = i^4 * i = 1 * i = i\ni ^6 = i^4 * i^2 = 1 * -1 = -1\n\netc.\n\nSo to solve any i ^n, just look at the remainder of n/4:\n\nremainder \nof n/4 ....... value of i ^n\n- - - -  - - - - - - - - - - - - - - -\n0 ....................... 1\n1 ........................ i\n2 ....................... -1\n3 ........................ -i\n\nSo now let's tackle your specific problem:\n\ni ^16 = 1 (since n=16, n/4 = 4 with remainder of 0)\ni ^6 = -1 (since n=6, n/4 = 1 with remainder of 2)\ni ^5 = i (since n=5, n/5 = 1 with remainder of 1)\ni ^13 = i (since n=13, n/4 = 3 with remainder of 1)\n\nSo, i^16 + i^6 - 2i^5 + i^13 = \n1 + (-1) - 2(i) + i =\n-i\n\nHope that helped!\n\n~ ♥ ~\n\nPS -- The "2-i" answers are incorrect.  Those answerers just made a tiny typo and took 1 - (-1) as the first two terms, rather than 1 + (-1).  Just wanted to let you know, so that you don't put the wrong answer down on your homework.
958498
. I - 20, 20, 18, 27, 36, 37, 45, 50\nC - 16, 18, 13, 21, 27, 26, 36, 39\n\n^ C ($ x 1000)\n- 40\n|.................... ...................... .................... ....................... •\n|.................... ...................... .................... ..................˚\n|\n|\n-30\n|\n|................... ....................... .................... .•\n|................... ....................... .................... ...˚\n|................... ....................... ......•\n-20\n|................... .................˚\n|................... .................˚\n|................... .............• \n|\n-10\n|\n|\n|\n| \n|________|________|____  ___|________|____  ___|__>\n................ 10.............. 20 ............. 30 .............40 ..............50 \n................................................. I ($ x 1000)\n\nThis is something like what your scattergram should look like\n\nThen you find a line using a linear regression on your calculator\n\nC = A + BI\n\nAnd then you put in 42 (for 42 000) and out comes your prediction for C in thousands of dollars
798483
Here it is:\n\n(98,42)Mo + (1,0)n ---> (99,42)Mo
374860
Well here is the math involved:\n\n(7i)^2 =  (7^2)(i^2) = 49(-1) = -49
683943
Any complex number can be expressed as r*e^(i*theta).\n\n2^i = r*e^(i*theta)\n\nlogs on both sides\n\nln(2^i) = ln (r) + ln (e^(i*theta))\ni*ln(2) = ln(r) + i*theta\n\nequate real and imag parts:\n\nln(r) = 0\nso\nr=1\n\ntheta = ln(2)\n\nso the answer is 2^i = e^(i*ln(2))\n\nThere's an identity to do that more easily, but I can't remember it - it's something like this:\n\na^b = fn(e^b)
817918
This is an oldie:  What's i^i, where i = sqrt(-1)?   Ans:  By Euler's formula, we know that e^pi i = -1.  Therefore e^(1/2) pi i = i, or e^-(1/2) pi = i^i = 0.20788.