what is an equivalent form of 2/(3+i)?

Question

explanation too, please. =)

A (3-i)/4
B (3-i)/5
C (4-i)/4
D (4-i)/5

Answer 1

multiply by (3-i)/(3-i) to get:

[2(3-i)] / [ 3^2 - i^2]

=2(3-i) / (9 +1)

= (3 - i) / 5

Answer 2

The correct answer is B. You need the multiply by what is called the complex conjugate. The complex conjugate has the same form as a complex expression with the sign reversed. (a+bi) has complex conjugate (a-bi) and vice versa.

So we have complex conjugate (3-i). If we place this top and bottom in a fraction we have (1). Let's multiply by our (1) to modify our expression. The top becomes 6-2i.
The bottom becomes 9 + 1 =10. The form of the result of multiplying complex congujates is always (a^2+b^2). In our case 3^2 +1^2. We can factor 2 out of everything to arrive at...
(3-i)/5

Answer 3

In this case, you see that all of the answers denote a real (or non-imaginery) denominator. The easiest way to change an imaginery number to a real number is to create a difference of squares, such as this (3+i) * (3-i) --> 9 - i^2.

In order to make certain the fraction is equal to its original state, you must multiply the numerator and the denominator by the same number. Therefore,

2/(3+i) --> 2(3-i) / (3+i)(3-i) --> 2(3-i) / [9 - i^2] --> 2(3-i) / 10 --> (3-i) / 5

Answer 4

2(3-i) / (3^2 - i^2) = 2(3-i) / 10
= (3-i) / 5
= B

Answer 5

You can multiply both the numerator and denominator by the conjugate (3 - i) of the denominator. By doing this, we are really just multiplying the entire fraction by 1, because x / x = 1 for all x. It just changes the way the number looks without changing its value.

[2 / (3 + i)] x [(3 - i) / (3 - i)] =
6 - 2i / [(9 - ( i²)] =
6 - 2i / [(9 - (-1)] =
6 - 2i / 10

We can factor a two out of both the numerator and denominator to get this final form:

(3 - i) / 5

So this last fraction is equivalent to the original, but looks different because of the conjugation process.

Answer 6

Ans: (3-i)/5

To arrive at this answer, you'll need to do some form of manipulation. We achieve this by multiplying both numerator and denominator of the term: 2/(3+i) by the denominator's "complex conjugate."

To obtain the complex conjugate, just change the sign between the real and imaginary parts (in other words, plus becomes minus and minus becomes plus). In this case, our divisor or denominator is (3+i), thus the complex conjugate is (3 - i).

1. Multiply both numerator and denominator by (3-i),
2. Use the relation: (i)^2 = -1 to simplify the product of denominator
3. Do further simplification of the resulting fraction to give the final answer.

Note that by multiplying the numerator and denominator by the same term means that you haven't changed it's value at all since it's essentially multiplying the original term by 1.

Answer 7

the answer is (3-i)/5
we need to rationalise the denominator to convert the base of the fraction in to a real number.the base of the fraction is a complex number (3+i).The rationalisin factor for it is (3-i).i.e, if u multiply (3+i) and (3-i) u'll get 10.
(3+i)*(3-i)=((3*3)+(3i)-(3i)-i^2).
i is an imaginary number and by the property of it i^2=-1

so the above eqn will be(9+1)=10

the fraction becomes
2(3-i)/(3+i)(3-i) which is equal to 2(3-i)/10=3-i/5

Answer 8

To get rid of the i in the denominator, multiply by potential of the conjugate: (3-i)/(3-i) 2/(3+i) x (3-i)/(3-i) =(6-2i)/10 (9-3i+3i-i^2) is the denominator...considering i^2=-a million, the denominator=10 =(3-i)/5 So the respond is B.

Answer 9

Complex conjugate of 3+i is 3-i

Multiply the numertor and denominator by 3-i to get

2(3-i)/(3^2 - i^2)
=2(3-i)/10
=(3-i)/5

B

Answer 10

[2/(3+i)][(3-i)/(3-i)] = 2(3-i)/(9-i^2) = 2(3-i)/(9+1) = (3-i)/5