Complex Numbers

Gerolamo Cardano is an Italian mathematician who introduced complex numbers. Complex numbers are those that can be expressed in the form of a+bi where a and b represent real numbers.

113-2i sqr 17

Well, there is doubtlessly more than one way to get from 9999 to 9, but the most direct way would be, 9999 - 9990 = 9.

The answer is relatively simple if you know hyperbolic functions.

Suppose x is real so that ix is an imaginary number.

Then tanh x = -i*tan(ix)

So tan(ix) = (tanh x )/-i = i*tanh x

= i * sinh x/csh x = i*(ex - e-x)/(ex + e-x) = i*(1 - e-2x)/(1 + e-2x)

It is 6+2i. But -6-2i will also serve.

You can, but the process is slightly complicated, because addition in the Complex field is like vector addition.

If z1 = (r1, a1), and If z2 = (r2, a2)

Then, if z = (r, a)

r = sqrt(r12 + r22)

and

a = arctan[(r1sina1 + r2sina2)/(r1cosa1 + r2cosa2)]

complex numbers with no real part
if any complex number z can be written a + i b
then pure imaginary numbers have a=0 and b not equal to 0

No. The square root of -24 isn't even real, let alone rational because the square root of any negative number is going to be an imaginary number.

That 5 is a factor.

It is: 360,000,000 or as 3.6*108 in scientific notation

No. A complex number consists of a real part and a imaginary part. If the real part equals zero, there is only the imaginary left and you could therefor argue that it is an imaginary number (or else it would still be a complex number -with a real part=0)

It is three million or 3.0*106 in scientific notation

2+6i

Zero is the additive identity in the set of real numbers; when you add zero to any number, the number does not change its identity.

Assume we have the product of these terms. Multiply these terms altogether to get 12i². Make note that i = √-1, i² = -1, i³ = -i and i⁴ = 1. Then, 12i² = 12(-1) = -12.

That is the answer to the question.

A [real] constant.

a variable

Complex numbers whose polar representation is (r, theta) where 3*pi/2 < theta < 2*pi.

a + i b, with a and b real numbers and i = sqrt(-1)

Assuming x is your number, x * 10^n = x moved n decimal places.

When n is positive, move the decimal point n places to the right.

When n is negative, move the decimal point n places to the left.

When n is 0, do nothing.

a more detailed solution of an answer

A diamond has two rotation symmetry. It is possible to have a diamond that does have four of rotation symmetry.

the mean or average of the set

arg(-2-i) = sqrt[22 + 12] = sqrt(5)

an imaginary number is imaginary so no (i guess) this answer kind of sucks