Question on complex numbers?
914511
x = (1 + 2j)(1 - 2j)\n= 1 - 2j + 2j - 4j^2 (j^2 means j squared and is how exponents are often written on computers)\n= 1 - 4j^2\nBut j^2 = -1, so\nx = 1 - 4j^2 = 1 - 4(-1) = 1 - (-4) = 1 + 4 = 5\n\n{You forgot that j was squared in going to your last step.  Also, you should probably write these as 4j rather than j4 to avoid confusion.  Note also that in mathematics, the letter i is used for √(-1).  j is often used by some engineers for this number, but you may find confusion in this forum and category by using j.}
454578
mrniceguy's answer is correct, but he has one little eror in the work.  i^7 = -i
572398
first multiply i by everything in the first bracket.\nso -8i * i + 3 * i + 2(2-5i)\n\nthen do same with 2nd bracket\nso -8i^2 + 3i + 2*2 - 5i*2\n= -8i^2 + 3i + 4 - 10i\n\nnow collect all the like terms\n-8i^2 + 3i - 10i + 4\n= -8i^2 - 7i + 4\n\nnow if i is the imaginary number i^2 would be -1.\nso it would be 8-7i +4 = 12-7i\n\nhope that helps. laterz
880047
I can find this on a TI-84+, is that how you want to get it or do you need an algebraic solution?
322093
Hello.  First you have to determine if you are going to find 'real' answers.  use b^2 - 4*a*c, where b=-3, a=1, c=5 (from ax^2 + bx + c).  \n\nNow, b^2 - 4*a*c = 3^2 - (4*1*5) = 9-20=-11, hence, you will have imaginary roots in your answer.\n\nNote that sqrt(-1) = i, then,\n\nx^2 - 3x + 5 = 0\n(x - [3/2])^2 + 5 - (3/2)^2 = 0 ;(Completing the roots)\n(x - [3/2])^2 = -11/4\n(x - [3/2])^2 = (i^2) * 11/4 ; (note that i^2 = -1)\n\nx - 3/2 = +/- i * 0.5 * sqrt(11)\nx = (3/2) +/- i * 0.5 * sqrt(11)\n\n------------\nAlways remember to check if b^2 - 4*a*c > 0 or < 0 then you will know if there is imaginary i in the answer..
752580
It just takes foiling-\n\n(5*2)+(5*5i)+(-3i*2) +(-3i*5i)\n(10)+(25i)+(-6i)+(-15*i^2)\ni^2=-1\nso...\n25+19i\n\nTo those who have said that you can't simplify it past 10 + 19i - 15i^2, that isn't correct considering that since i is the square root of -1, i^2 is -1. That's why you can simplify -15i^2 to 15, thus making it 10+19i+15, or 25+19i
1085635
First of all, you (along with some other people here, seem to be confusing imaginary and complex numbers).  Complex numbers make a big set.  Imaginary numbers make a smaller set inside the complex numbers and real numbers make a smaller set inside the complex numbers.  These two sets, however, do not make the entire set of complex numbers.  In other words,\n1.All real numbers are complex numbers.\n2.All imaginary numbers are complex numbers.\n3.There are complex numbers that are not real number and they are not imaginary numbers either.\n\nDraw a big circle and draw two smaller circles inside the big circle.\n\nSo from now on, everybody, be careful when you use the words imaginary and complex.  They are not the same things.\n\nNow for some clarifications:\n1.A complex number can always be written as a+bi where i is the imaginary unit (square root of -1).\n2.In any complex number a+bi, a is called the real part.\n3.And b is called the imaginary part.\n\nSo the number 5 for example, is a real number AND it is also complex because 5 can be written as 5=5+0i.\n\nThe number 2i for example, is an imaginary number AND it is also a complex number because 2i=0+2i.\n\nSo now the light bulb goes off and everyone realizes that a real number is just a complex number with the imaginary part being zero and an imaginary number is a complex number with the real part being zero.\n\nThe number 1+2i is the last example.  It is a number that is not real not imaginary.\n\nHere is an interesting thing.  On the real numbers, we have this notion of ordering, meaning if I give you any two real numbers, you can always put <, >, or = between them, meaning you can always compare them.  The real numbers are ordered, in fact, they are "well-ordered".  What about the complex numbers that?  How in the world do you compare 1+i to 1-i?  Is it bigger than, less than, or even equal that goes between them?  Well, the answer is that they are both equal to each other.  But dig in, and see if you can find out why.\n\n----------------------------------------\nNow for the question you asked,  pretty much all the engineering fields make extensive use of complex numbers but electrical engineers live, eat, drink, and breathe them.\n\nBut no worries, they were not invented only for them.  In math, if you go far enough, it turns out that it actually is a lot easier to work with complex numbers than with real numbers only.\n\nHere is an example you should be familiar with, when we want to find roots of polynomials.\nIf we consider a polynomial, as a polynomial with real numbers, then we know for a fact that you can't always find all of the zeros.  You might not find any roots at all. x^2+1=0 has all real coefficients but no roots which is a problem.\nNow, if we consider the same polynomial as a polynomial with complex numbers, we are ALWAYS guaranteed ALL the roots.  Because remember, all real numbers are complex numbers as well. So x^2+1=0 is a polynomial with complex coefficients.
157955
9*9\n=(3^2)(3^2)\n=3^(2+2)\n=3^(4)\n=3*3*3*3\n=(3*3)*(3*3)\n=(9)(9)\n=1(9*9)\n=1(9+9+9+9+9+9+9+9+9)\n=1(18+18+18+18+9)\n=1(36+36+9)\n=1(72+9)\n=1(81)\n=1*81\n=81\nFINALLY, I GOT IT!\nThe answer to this complex rational algebraic problem is 81\n\nDuh!  ;-p
1060909
Let z=x+iy\n\nThen |Re(z)|=|x|\n|Im(z)|=|y|\n\nAnd |z|=sqrt(x^2+y^2), by definition.\n\nSo now you seek to prove:\n|x|+|y|<= sqrt(2)sqrt(x^2+y^2)=sqrt(2x^2+2y^2)\n\nSince |x|>0 and |y|>0, we can square both sides:\n\nx^2 +2|xy|+y^2<=2x^2+2y^2\n\nThis is true if and only if\n\nx^2-2|xy|+y^2>=0, which is true iff\n\n(|x|-|y|)^2>=0. This however is necessarily true, since x and y are real numbers, and the square of a real number is always positive. Since this statement is true, the original statement \n\n|Re(z)|+|IM(z)| =< sqrt 2 * |z|\n\nis also necessarily true.
1003676
You have a theorem leads to :\nce * ed = ae * eb\n6 * 5 = ( 2 - eb)*eb\n30 = 2 eb - (eb)^2\n(eb)^2 - 2 eb + 30 = 0\nWhich have imaginary roots\nif ae = 2   ( not ad as you mentioned )\nthen 30 = 2*eb\ntherefor eb = 15
673741
just multiply it\n\n-35-45i-14i+18i^2\n\nthen in complex no u know that value of i^2=-1\n-35-59i+18*(-1)\n-18-35-59i\n-53-59i
184348
if 'i' means the imaginary unit....\ni = e^(i*pi/2) = cos(pi/2) + i*sin(pi/2)\nsquare root of i\n(i)^(1/2) = e^(i*pi/4) = cos(pi/4) + i*sin(pi/4)\ncube root of i\n(i)^(1/3) = e^(i*pi/6) = cos(pi/6) + i*sin(pi/6)\nand so on.......
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
770326
If you are allowed to head into the realm of complex/imaginary numbers, you could factor out a sqrt(-1) and replace it with a lowercase i.\n\nsqrt(-24) = sqrt(4) x sqrt(6) x sqrt(-1)\nsqrt(-24) = ± 2 sqrt(6) i
1038170
for complex numbers a , b exponentiation is defined as\na^b = e^(b*Log(a))\nWhere Log(a) is defined using the modulus r and argument t of a:\nLog(a)=ln(r)+it\nHere t can take infinitely many values. It may differ by a multiple of 2*Pi without changing the a.\n\nReturning to main question:\nFirst we note that Log(i) has infinitely values as folows:\nLog(i)=i(4k+1)Pi/2 where k is an integer.\nNow\ni^e = e^(e*Log(i)) = e^(i*Pi*e*(4k+1)/2)\n\ntaking a(k)=Pi*e*(4k+1)/2\nwe have\ni^e = e^ia(k) = cos(a(k))+i*sin(a(k))\n(infinitely many values)
731516
The standard form of a complex number is a + bi.  Also, we can't leave any radicals in the denominator of a fraction, so we will have to rationalize the denominator.  Since this denominator has 2 terms, we will use the conjugate (same terms but different sign in the middle).  The conjugate of 3-4i is 3+4i.  So, multiply the fraction by (3+4i)/(3+4i).  In the denominator, you can use "FOIL" to multiply -- remember that i^2=(-1).  You should end up with (3-4i)/25, which you can then write as 3/25 + 4i/25.
1071898
it's x, assuming that x is zero or postive. \n\nif x is negative, it is either undefined, \nor if you use complex numbers, it is |x|=-x.