Question

Plsss give me an example how to do it
Am deaddd

Answer 1

(1) (i)

7i(3+5i) = 21i + 35i²

= 21i +35(-1)

= -35+21i

because iota(i) = √(-1), i²= -1 and so on

2 (iii)

in (a+bi) ,a is real part and b imaginary part with iota

(6+i)(-2+3i)= -12-2i+18i+3i²

= -12 +16 i +3(-1)

= -12+16i -3

= -15 + 16i

3(iv)

(-2-2i)(-2+2i)= 4+4i-4i-4i²

= 4-4i²= 4 - 4(-1)= 4+4 = 8

= 8 + 0i

4 sum is same as above

5(i)

(√3+2i)(√3-2i) = (√3)² - (2i)²

= 3-4i² = 3 - 4(-1) = 3+4 = 7

= 7+ 0i