Fine the real and imaginary part of 6i/(2-i)²

6 i/(2-i) * 1/(2-i)

6i (2+i)/ [ (2-i)(2+i) ] * (2+i)/[ (2-i)(2+i) ]

(12 i -6) /[ 4+1 ] * (2+i)/ [ 4+1 ]

(24 i-12-12 -6 i)/25

-24/25 + (18 /25) i

Please which is the real and Aldo the imaginary

Hey, you better take the course

if you have
x + y i
x is real part
y i is imaginary part
see:
https://www.mathsisfun.com/numbers/complex-numbers.html

Thanks friends. I really appreciate . now I understand

To find the real and imaginary parts of the expression (6i/(2-i)²), we need to simplify the expression and then split it into real and imaginary parts.

Let's start by simplifying the expression (2-i)²:

(2-i)² = (2-i)(2-i)
= 2(2) - 2(i) - i(2) + i(i)
= 4 - 2i - 2i + i²
= 4 - 4i + i²

Note that i² is equal to -1:

(2-i)² = 4 - 4i + (-1)
= 3 - 4i

Now, we can substitute this simplified expression back into our original expression:

6i/(2-i)² = 6i/(3 - 4i)

To simplify further, we multiply the numerator and the denominator by the conjugate of the denominator:

6i/(3 - 4i) * (3 + 4i)/(3 + 4i)
= (6i * (3 + 4i)) / ((3 - 4i) * (3 + 4i))
= (18i + 24i²) / (9 + 12i - 12i - 16i²)
= (18i + 24(-1)) / (9 + 12(-1) - 12(-1) - 16(-1))
= (18i - 24) / (25)
= (18i / 25) - (24 / 25)

Now we have split the expression into real and imaginary parts:

Real part: -24 / 25
Imaginary part: 18i / 25

So, the real part is -24/25 and the imaginary part is 18i/25.