Product of Imaginary Numbers

let the two img. number is ia and ib where a and b are two real number. Note that numbers are img. not complex. Now product of two number is (ia)(ib)=ab(i^2). Since i^2=-1 hence we have product as -ab which is a real number. So Product of two img. number is always a real number.

The product of two imaginary numbers does not necessarily is a rational or irrational number, but it's always be a real number.

Imaginary number contains " i " in it so after multiplication we`ll get i^2 = -1 a real number

real and negative

For eg take two imaginary numbers as i and 2i and the product of these two are -2 which is a real number.

Consider the first imaginary number we learned in school, i, the square root of of negative one. If we square it, we get a real, rational number. So if the product of two imaginary numbers is ALWAYS something, it has to be one of these, since i^2 is one of them. Then consider if you can add a term to one of the i's to make the solution no longer either rational or real. If we multiply sqrt2i x i, then we get a real irrational number. Presuming that the questions offers us a correct answer, it has to be the other one. There is nothing you can add to i such that when you square it you get an imaginary number.

real number..........i.e ai bi=-ab (i i=-1) which is real, but for complex number it is imaginary too

Suppose two imaginary numbers 2i and 3i . Then their product is (2i)(3i) which is equal to 6i^2 as we know i^2=-1 so it is -6 which is a real number ......

Let two purely imaginary nos. are z1=ai and z2=bi ; a,b in R, then z1*z2 = ab(i^2)=-ab in R; as by definition i^2 =-1.

take the example 3i and 4i both are imaginary number now product of these both number is 12i^2 since i^2=-1 then ans will be -12 is puerly real

The Product of 2 Imaginary units i.e 'i' is always a negative real number. i * i = -1. Eg : Let 3i and 4i be two imaginary numbers. The product is 12 (i)^2. Now, the value of i^2 will decide whether the result is real or imaginary. Here's the catch : According to the standards of Mathematics, the square of the imaginary unit 'i' is always equal to '-1'. Therefore, we have the solution as '12 (-1) = -12; which thus is a real number. You can test it with different values of imaginary numbers and get the same analysis. Also remember that a 'Complex' number is different from an Imaginary number.Few people multiply complex numbers and show complex numbers as results and say that the result is not always real. :P

let us consider any two imaginary number i^n and i^m. then product is i^m+n you can put arbitrary values of m and n .we can get areal number. e.g...m=1 , n=1 then product is i^2=-1 which is real number.

if i=square root of -1 then root x root will "cancel" each other, it remains -1, which is a real number

i^2=i.i=-1=real number

2i*2i=4i^2=-4 where i^2=-1