Complex Numbers

Euler's formula states that eix = cos(x) + isin(x) where i is the imaginary number and x is any real number.

First, we get the power series of eix using the formula:

ez = Σ∞n=o zn/n! where z = ix. That gives us:

1 + ix + (ix)2/2! + (ix)3/3! + (ix)4/4! + (ix)5/5! + (ix)6/6! + (ix)7/7! + (ix)8/8! + ...

which from the properties of i equals:

1 + ix - x2/2! - ix3/3! + x4/4! + ix5/5! - x6/6! - ix7/7! + x8/8! + ...

which equals:

(1 - x2/2! + x4/4! - x6/6! + x8/8! - ...) + i(x - x3/3! + x5/5! - x7/7! + ...).

These two expressions are equivalent to the Taylor series of cos(x) and sin(x). So, plugging those functions into the expression gives cos(x) + isin(x).

Q.E.D.

Of course, were you to make x = п in Euler's formula, you'd get Euler's identity:

eiπ = -1

The answer above assumes that the exponential function is defined, for all complex numbers, by its power series. (Or, at least, that someone else has already proved that the power series definition is equivalent to whatever we're taking to be the official definition).

So the answer depends critically on what the definition of the exponential function for complex numbers is.

Suppose you know everything about the real numbers, and you're trying to build up a theory of complex numbers. Most people probably view it this way: Mathematical objects such as complex numbers are out there somewhere, and we have to find them and work out their properties. But mathematicians look at it slightly differently. If we can construct a thing which has all the properties we feel the field of complex numbers should have, then for all practical purposes the field of complex numbers exists. If we can define a function on that field which has all the properties we think exponentiation on the complex numbers should have, then that's as good as proving that complex numbers have exponentials. Mathematicians are comfortable with weird things like complex numbers, not because they have proved that they exist as such, but because they have proved that their existence is consistent with everything else. (Unless everything else is inconsistent, which would be a real pain but is very unlikely.)

So how do we construct this exponential function? One approach would be simply to define it by exp(x+iy) = ex(cos(y)+i.sin(y)). Then the answer to this question would be trivial: exp(iy) = exp(0+iy) = e0(cos(y)+i.sin(y)) = cos(y)+.sin(y). But you'd still have some work to do to prove things like exp(z+w) = exp(z).exp(w). Alternatively, you could define the exponential function by its power series. (There's a theorem that lets you calculate the radius of convergence, and that tells us the radius is infinite, i.e. the power series works everywhere.) Or maybe you could try something like proving that the equation dw/dz = w has a unique solution up to a multiplicative constant, and defining the exponential function to be the solution which satisfies w=1 at z=0.

Let z = cos(x) + i*sin(x)

dz/dx = -sin(x) + i*cos(x)

= i*(i*sin(x) + cos(x))

= i*(cos(x) + i*sin(x))

= i*(z)

therefore dz/dx = iz

(1/z)dz = i dx

INT((1/z)dz = INT(i dx)

ln(z) = ix+c

z = e(ix+c)

Substituting x=0, we get cos(0) + i*sin(0) = e(0i+c)

1+0 = e(0+c)

e(0+c) = 1

therefore c=0

z = eix

eix = cos(x) + i*sin(x)

Substituting x=pi, we get e(i*pi) = -1 + 0

e(i*pi) = -1

e(i*pi) + 1 = 0

qED

The real number system is a mathematical field.

To start with, the Real number system is a Group. This means that it is a set of elements (numbers) with a binary operation (addition) that combines any two elements in the set to form a third element which is also in the set. The Group satisfies four axioms: closure, associativity, identity and invertibility.

Closure: For all x and y in the set, x+y is also in the set.

Associativity: For all x, y and z in the set, (x+y)+z = x+(y+z)

Identity: there exists an element, normally denoted by 0, such that for any element x in the set, 0+x = x = x+0.

Invertibility: For every element x in the set, there is an element y in the set such that x+y = 0 = y+x. y is usually written as -x.

In addition, it is a Ring. A ring is an Abelian group (that is, addition is commutative) and it has a second binary operation (multiplication) that is defined on its elements. This second operation is distributive over the first.

Abelian (commutative): For every x, y in the set, x+y = y+x.

Distributivity of multiplication over addition: For any x, y and z in the set, x*(y+z) = x*y + x*z.

And finally, a Field is a Ring over which division - by non-zero numbers - is defined.

The four axioms of a Group are satisfied for the second operation. The multiplicative identity is denoted by 1, the multiplicative inverse of an element x is denoted by 1/x or x^-1.

To solve:

x+y = 4

x*y = 13

x=4-y, so

(4-y)*y = 13 iff

y^2 - 4y + 13 = 0 iff

(y-2)^2 = -9 iff

y-2 = +- 3i

y = 2 +- 3i

e.g., x = 2+3i, y = 2-3i

Not sure what you're asking, but being the question is in Complex Numbers category, it could be this:

• If the real part of a complex number is zero, then it is a pure imaginary number.

Try this

input "z =?";x,y

print "z=";x;"+";y;"i"

input "w =?";u,v

print "w=";u;"+";v;"i"

print "z+w = ";x+u;"+";y+v;"i"

print "z*w = ";x*u-y*v;"+";x*v+y*u;"i"

example output:

z=? 1,2

w=? 2,3

z+w = 3+5i

z*w = -4+7i

Note that it is possible to input negative imaginary parts. To display the sum and product correctly in that case, you need to make an intermediate step in the program which assigns the imaginary part to a separate variable, decide whether or not the imaginary part is negative, and then print a "+" if it is positive or no sign when it is negative (printing a negative number will automatically print the negative sign).

Of course, you need to modify the program to your computer language.