could u check my imaginary word problem please?its math?

Question

(-2+3i) ; (3+5i)

The problem above needs to be solved in addition, subtraction and multiplication, and division and needs to be in standard form.

Heres my answer: addition= 1+8i
Multiplication= -6+14i
subtract= -5-2i
division= could someone help me out for this one

Answer 1

Add : 1 + 8i
Subtract : -5 - 2i
Multiply : -21 - i
Divide : (34 + 19i)/34

Multiplying complex numbers
i^2 = -1

(-2 + 3i)(3+5i)
= -6 + 9i - 10i + 15i^2
= -6 - i - 15
= -21 - i

Dividing complex numbers :
You multiply the numerator and denominator by the complex conjugate of the denominator.
eg.
(a + bi) * (c - di)
--------
(c + di) * (c - di)

This has the effect of turning the denominator into a real number.

(-2 + 3i) * (3 - 5i)
--------------------- =
(3 + 5i) * (3 - 5i)

-6 + 10i + 9i - 15i^2
------------------------- =
9 - 15i + 15i - 25i^2

9 + 19i
------------
34

Answer 2

(-2 + 3i) ; (3 + 5i)
(-2 + 3i) + (3 + 5i) = 1 + 8i

(-2 + 3i) - (3 + 5i) = -5 + 2i

(-2 + 3i)(3 + 5i) = -6 - 10i + 9i - 15 = -21 - i

To divide complex numbers, multiply both numerator and denominator by the complex conjugate of the denominator. This removes the complex term in the denominator.
(-2 + 3i) / (3 + 5i) =
(-2 + 3i)(3 - 5i) / (3 + 5i)(3 - 5i) =
(-6 + 10i + 9i + 15) / (9 + 25) =
(9 + 19i) / 34

Answer 3

(-2+3i)(3+5i)= -6-10i+9i-15
=-21-i
note i squared = -1

(-2+3i)(3-5i)/(3+5i)(3-5i)=(-6+10i+9i+15)/(9+25)

note realise the denominator

=(9+19i)/34

Answer 4

I don't think your multiplication is correct:

(-2 + 3i) (3 + 5i)
FOIL
-6 - 10i + 9i +15i²
-6 -i -15
-21-i

I don't remember how to divide complex numbers. sorry.

Answer 5

Kookiemon has it right, I was rather slow, not having done any complex arithmetic for about 20 years.

The following is an excellent mathematics reference site:
http://mathworld.wolfram.com/