Express in the form of a+ib

(3+i√5)(3-i√5)÷(√3+√2i)-(√3-i√2)

If your expression mean:

( 3 + i √5 ) ( 3 - i √5 ) / [ ( √3 + √2 i ) - ( √3 - i √2 ) ]

then:

[ ( 3 + i √5 ) ∙ ( 3 - i √5 ) ] / [ ( √3 + √2 i ) - ( √3 - i √2 ) ] =

[ 3 ∙ 3 + 3 ∙ i √5 - i √5 ∙ 3 - i √5 ∙ i √5 ] / [ √3 + √2 i - √3 - ( - √2 i ) ] =

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

( 9 - i² ∙ √5² ) / [ √3 - √3 + √2 i + √2 i ] =

( 9 - ( - 1 ) ∙ 5 ) / ( √2 i + √2 i ) =

[ 9 - ( - 5 ) ] / 2 √2 i =

( 9 + 5 ) / 2 √2 i =

14 / 2 √2 i =

2 ∙ 7 / 2 √2 i =

7 / √2 i =

7 ∙ i / √2 i ∙ i =

7 i / √2 i² =

7 i / √2 ∙ ( - 1 ) =

7 i / - √2 =

- 7 i / √2 =

0 - 7 i / √2 =

0 - i 7 / √2

( 3 + i √5 ) ( 3 - i √5 ) / [ ( √3 + √2 i ) - ( √3 - i √2 ) ] = 0 - i 7 / √2

Can you see the confusion created by insufficient use of brackets ?

I will take it as face value the way you typed it ....

(3+i√5)(3-i√5)÷(√3+√2i)-(√3-i√2) ------> why is i in front of -i√2, but behind in √2i ??
= (9 - 5i^2) ÷ (√3+i√2) - (√3-i√2)
= 14/(√3+i√2) - (√3-i√2)
= 14/(√3+i√2)*( (√3-i√2)/(√3-i√2) ) - (√3-i√2)
= 14(√3 - i√2)/5 - (√3 - i√2)
= 14(√3 - i√2)/5 - 5(√3 - i√2)/5
= (9/5)(√3 - i√2)

To express the given expression in the form of a+ib, let's simplify step by step:

Step 1: Simplify the numerator (3+i√5)(3-i√5):
Using the formula (a+b)(a-b) = a^2 - b^2, we can simplify it as follows:
(3+i√5)(3-i√5) = 3^2 - (i√5)^2 = 9 - (-5) = 9 + 5 = 14

Step 2: Simplify the denominator (√3+√2i)-(√3-i√2):
To simplify this expression, let's simplify each part separately:
a) (√3 + √2i):
This is already in the desired form, as it contains both a real part (√3) and an imaginary part (√2i).

b) (√3 - i√2):
To simplify this part, let's multiply the terms by the conjugate of the denominator (√3 + i√2):
(√3 - i√2) * (√3 + i√2) = √3 * √3 + √3 * i√2 - i√2 * √3 - i√2 * i√2
= 3 + i√6 - i√6 - i^2 * 2
= 3 + i√6 - i√6 + 2
= 5

So, (√3 - i√2) simplifies to 5.

Step 3: Combine the numerator and denominator:
(3+i√5)(3-i√5) ÷ (√3+√2i) - (√3-i√2) = 14 ÷ (√3 + √2i) - 5

Step 4: Rationalize the denominator (√3 + √2i):
To rationalize the denominator, we need to multiply the numerator and denominator by the conjugate of (√3 + √2i), which is (√3 - √2i):
(14 ÷ (√3 + √2i)) * ((√3 - √2i) / (√3 - √2i)) = (14 * (√3 - √2i)) / ((√3)^2 - (√2i)^2)
= (14 * (√3 - √2i)) / (3 - (-2i^2))
= (14 * (√3 - √2i)) / (3 + 2)
= (14 * (√3 - √2i)) / 5

Finally, the expression can be expressed as (14 * (√3 - √2i)) / 5.

Note: Depending on the specific context, it may be required to further simplify or write the expression differently.