Is the sum of two imaginary numbers always an imaginary number? Justify your answer.?

Question

Is the productt of two imaginary numbers always an imaginary number? Justify your answer.

Answer 1

The sum of two imaginary numbers is not always imaginary.
Let's take ai, an imaginary number, and -ai another imaginary number. The sum is 0 (zero) which is NOT an imaginary number.

The product of two imaginary numbers is NOT an imaginary number. If I have "ai" and "bi" and multiply them, I will have abi^2, and since i^2 is -1, the result is real number "-ab".

Answer 2

Regards,

By imaginary number do you mean complex number?

If so then read on...

Yes, always, the set of complex numbers is closed under addition and multiplication.

let z1 = a + bi where a and b are real numbers

let z2 = c + di where b and d are real numbers

then z1 + z2 = (a+c) + (b+d)i

a+c and b+d are both real numbers

so z1 + z2 is a complex number.

eg 0 = 0 + 0i

similarly z1 * z2 = (a + bi ) *(c + di) = (ac -bd) + (ad + bc)i

ac - bd is real as is ad + bc therefore z1*z2 is complex.

Now each complex number has a real and imaginay part.

The portion with the i is the imaginary part.

So it is possible to have complex numbers which are entirely imaginary or real.

Regards

Answer 3

An imaginary number is represented as ai where a is any positive or negative number, and i is a special constant whose square equals -1 (This means i*i = -1).

So, let's take two imaginary numbers : ai and bi
So, the sum is:
ai + bi = (a+b)i
The sum is an imaginary number
Their product is:
ai * bi = (a*b)*(i*i) = (a*b)*(-1) = -a*b = -ab
Their product is a real number.

Answer 4

an imaginary number is a number times i (the square root of negative1)
if they are added: 5i + 3i = 8i,or subtracted 2i - 9i = -7i
then they will always be imaginary

square root -1 = i
so
i*i = -1
so
if multiplyed: 4i*3i = -12
a real number, as the two i's are changed back into -1, then the two numbers are multiplied like:
(i*i)*(4*3) = (-1)*(12) = -12

Answer 5

Provided you're referring to complex numbers, i.e. complex numbers of the form (x + yi) for real numbers x and y, the answer is yes. The proof uses the fact that the sum of real numbers is a real number.

Let x = a + bi,
y = c + di
for some real numbers a, b, c, d.

Then
x + y = (a + bi) + (c + di)
= (a + c) + (b + d)i

Which shows that the sum of two complex numbers is also a complex number, since (a + c) is a real number and (b + d) is a real number.

Similarly, the product of two complex numbers is also a complex number. Consider:

xy = (a + bi)(c + di)
= ac + adi + bci + bd i^2

But remember that i^2 = -1, so we get

= ac + (ad + bc)i + bd(-1)
= ac + (ad + bc)i - bd
= ac - bd + (ad + bc)i

As you can see, (ac - bd) is a real number, (ad + bc) is a real number, showing the product of two complex numbers is itself complex.

Answer 6

Yes.
If positive+positive=positive,
and
negative+negative=negative,
then
imaginary+imaginary=imaginary!

Answer 7

What are you on?
Imaginary numbers. -_-

Answer 8

( a + i b ) + (c - i b)
(a + c) which is not imaginary.