question: Why does the square root of i = (1/square root of 2)(1+i)?
comment: Complex numbers have a real/imaginary part. So 'i' really = 0 + 1i then to find square root: * (a + bi)^2 = 0 + 1i * a^2 + 2abi + b^2 * i^2 = 0 + 1i : But i^2 == -1 * a^2 + 2abi - b^2 = 0 + 1i : Separate real/imaginary parts * (a^2 - b^2 ) = 0 : Means a == b * 2abi = 1i = > 2aai = 1i * 2a^2 i = 1i : divide by i * 2a^2 = 1 * a^2 = 1/2 * a = sqrt(1/2) = 0.707106... plug back into original (a + bi) formula you get 0.707 + 0.707i
passages: In a field of characteristic 2, an element either has one square root or does not have any at all, because each element is its own additive inverse, so that . If the field is finite of characteristic 2 then every element has a unique square root. In a field of any other characteristic, any non-zero element either has two square roots, as explained above, or does not have any.
   Because of the discontinuous nature of the square root function in the complex plane, the following laws are not true in general.
  
  BULLET::::- formula_23 (counterexample for the principal square root: and ) This equality is valid only when formula_24
  
  BULLET::::- formula_25 (counterexample for the principal square root: and )This equality is valid only when formula_26
  
  BULLET::::- formula_27 (counterexample for the principal square root: )This equality is valid only when formula_28
  
  A similar problem appears with other complex functions with branch cuts, e.g., the complex logarithm and the relations or which are not true in general.
   Each element of an integral domain has no more than 2 square roots. The difference of two squares identity is proved using the commutativity of multiplication. If and are square roots of the same element, then . Because there are no zero divisors this implies or , where the latter means that two roots are additive inverses of each other. In other words if an element a square root of an element exists, then the only square roots of are and . The only square root of 0 in an integral domain is 0 itself.
   A square root can be constructed with a compass and straightedge. In his Elements, Euclid (fl. 300 BC) gave the construction of the geometric mean of two quantities in two different places: Proposition II.14 and Proposition VI.13. Since the geometric mean of "a" and "b" is , one can construct simply by taking .
  
  The construction is also given by Descartes in his "La Géométrie", see figure 2 on page 2. However, Descartes made no claim to originality and his audience would have been quite familiar with Euclid.
   Section::::Complex roots.
  
  Every complex number other than 0 has "n" different "n"th roots.
  
  Section::::Complex roots.:Square roots.
  
  The two square roots of a complex number are always negatives of each other. For example, the square roots of are and , and the square roots of are
  
  If we express a complex number in polar form, then the square root can be obtained by taking the square root of the radius and halving the angle:
  
  A "principal" root of a complex number may be chosen in various ways, for example
   Of course, there are other solutions, which is evidenced by considering the position of i in the complex plane and in particular its argument arg "i". We can rotate counterclockwise π/2 radians from 1 to reach i initially, but if we rotate further another 2π we reach i again. So, we can conclude that i(π/2 + 2π) is "also" a solution for log i. It becomes clear that we can add any multiple of 2πi to our initial solution to obtain all values for log i.
   Thus, if and are coprime, is also a primitive th root of unity, and therefore there are (where is Euler's totient function) distinct primitive th roots of unity. (This implies that if is a prime number, all the roots except are primitive.)
  
  In other words, if is the set of all th roots of unity and is the set of primitive ones, is a disjoint union of the :
  
  where the notation means that goes through all the divisors of , including and .
  
  Since the cardinality of is , and that of is , this demonstrates the classical formula
   where is the largest integer smaller than . Note that both the numerator and the denominator have been multiplied by a number smaller than 1.
  
  Through this, and by multiplying out both the numerator and the denominator, we get:
  
  It follows that can be replaced with :
  
  Then, can also be replaced with in the denominator:
  
  The square of can be replaced by 3. As is multiplied by , their product equals :
   Thus finding "n"th roots in the complex plane can be segmented into two steps. First, the magnitude of all the "n"th roots is the "n"th root of the magnitude of the original number. Second, the angle between the positive horizontal axis and a ray from the origin to one of the "n"th roots is formula_102, where formula_98 is the angle defined in the same way for the number whose root is being taken. Furthermore, all "n" of the "n"th roots are at equally spaced angles from each other.
   Gauss proved that a primitive th root of unity can be expressed using only square roots, addition, subtraction, multiplication and division if and only if it is possible to construct with compass and straightedge the regular -gon. This is the case if and only if is either a power of two or the product of a power of two and Fermat primes that are all different. 
   Wrongly assuming one of these laws underlies several faulty "proofs", for instance the following one showing that :
  
  The third equality cannot be justified (see invalid proof). It can be made to hold by changing the meaning of √ so that this no longer represents the principal square root (see above) but selects a branch for the square root that contains ()·(). The left-hand side becomes either
  
  if the branch includes +"i" or
  
  if the branch includes −"i", while the right-hand side becomes
   BULLET::::- For , the cyclotomic polynomial is Therefore, the only primitive first root of unity is 1, which is a non-primitive th root of unity for every "n" greater than 1.
  
  BULLET::::- As , the only primitive second (square) root of unity is –1, which is also a non-primitive th root of unity for every even . With the preceding case, this completes the list of real roots of unity.
  
  BULLET::::- As , the primitive third (cube) roots of unity, which are the roots of this quadratic polynomial, are
   BULLET::::- formula_2, since a ground term of height up to "h"+1 can be obtained by composing any "i" ground terms of height up to "h", using an "i"-ary root function symbol. The sum has a finite value if there is only a finite number of constants and function symbols, which is usually the case.
  
  Section::::Formal definition.:Building formulas from terms.
   In the section Elementary properties, it was shown that if is the set of all th roots of unity and is the set of primitive ones, is a disjoint union of the :
  
  This implies
  
  Applying the Möbius inversion formula gives
  
  In this formula, if , then , and for : . Therefore, .
  
  This is the special case of Ramanujan's sum , defined as the sum of the th powers of the primitive th roots of unity:
  
  Section::::Orthogonality.
  
  From the summation formula follows an orthogonality relationship: for and 
   This means that in polar coordinates, we are taking the cube root of the radius and dividing the polar angle by three in order to define a cube root. With this definition, the principal cube root of a negative number is a complex number, and for instance will not be −2, but rather .
  
  This difficulty can also be solved by considering the cube root as a multivalued function: if we write the original complex number "x" in three equivalent forms, namely
  
  The principal complex cube roots of these three forms are then respectively
   This is also true for negative exponents. In particular, the reciprocal of an th root of unity is its complex conjugate, and is also an th root of unity:
  
  If is an th root of unity and then . In fact, by the definition of congruence, for some integer , and
  
  Therefore, given a power of , one has , where is the remainder of the Euclidean division of by .
   Another way to find the square root of an "n" × "n" matrix "A" is the Denman–Beavers square root iteration. 
  
  Let "Y" = "A" and "Z" = "I", where "I" is the "n" × "n" identity matrix. The iteration is defined by
  
  As this uses a pair of sequences of matrix inverses whose later elements change comparatively little, only the first elements have a high computational cost since the remainder can be computed from earlier elements with only a few passes of a variant of Newton's method for computing inverses,
   The cosine of the angle between two roots is constrained to be a half-integral multiple of a square root of an integer. This is because formula_36 and formula_37 are both integers, by assumption, and
   Using a hyperbolic ruler, construct AI' such that AI' is parallel to "a"' and perpendicular to "a." Construct another line A'I such that A'I is parallel to "a" and perpendicular to "a'." Let the intersection of AI' and A'I be B. Because ᗉ IBI'  /2 radians"," the case now plays out like case 1, allowing construction of the common parallel to BI and BI'.
  
  "Case 3: a and a' are ultraparallel to each other"
   Unlike in an integral domain, a square root in an arbitrary (unital) ring need not be unique up to sign. For example, in the ring formula_33 of integers modulo 8 (which is commutative, but has zero divisors), the element 1 has four distinct square roots: ±1 and ±3.
  
  Another example is provided by the ring of quaternions formula_34 which has no zero divisors, but is not commutative. Here, the element −1 has infinitely many square roots, including , , and . In fact, the set of square roots of −1 is exactly
   where formula_3 are real or complex numbers and formula_4; thus is the degree of the polynomial.
  
  Section::::Continuous dependence on coefficients.
  
  The roots of a polynomial of degree depend continuously on the coefficients. For simple roots, this results immediately from the implicit function theorem. This is true also for multiple roots, but some care is needed for the proof.
   If is a primitive th root of unity, the same is true for , and formula_36 is twice the real part of . In other words, is a reciprocal polynomial, the polynomial formula_37 that has as a root may be deduced from by the standard manipulation on reciprocal polynomials, and the primitive th roots of unity may be deduced from the roots of formula_37 by solving the quadratic equation formula_39 That is, the real part of the primitive root is formula_40 and its imaginary part is formula_41
   For four other values of , the primitive roots of unity are not quadratic integers, but the sum of any root of unity with its complex conjugate (also an th root of unity) is a quadratic integer.
  
  For , none of the non-real roots of unity (which satisfy a quartic equation) is a quadratic integer, but the sum of each root with its complex conjugate (also a 5th root of unity) is an element of the ring Z[] (). For two pairs of non-real 5th roots of unity these sums are inverse golden ratio and minus golden ratio.
   BULLET::::6. It follows that must be even (as squares of odd integers are never even).
  
  BULLET::::7. Because is even, there exists an integer that fulfills: .
  
  BULLET::::8. Substituting from step 7 for in the second equation of step 4: is equivalent to , which is equivalent to .
  
  BULLET::::9. Because is divisible by two and therefore even, and because , it follows that is also even which means that is even.
  
  BULLET::::10. By steps 5 and 8 and are both even, which contradicts that is irreducible as stated in step 3.
   The error here lies in the last equality, where we are ignoring the other fourth roots of 1, which are −1, "i" and −"i" (where "i" is the imaginary unit). Since we have squared our figure and then taken roots, we cannot always assume that all the roots will be correct. So the correct fourth roots are "i" and −"i", which are the imaginary numbers defined to square to −1.
  
  Section::::Power and root.:Complex exponents.
  