please can anyone solve this question and show workings.
(i/5)-((2-2i)/(4-3i)).
Thank you

Which kind maths question be this?

it's a complex number question

what's 2-2i?

it's a complex number, 2 is the real part while 2i is the imaginary part.
Have you any idea of what a complex number is?

Here a tutorial. I'm kinda confused.

http://www.nipissingu.ca/algebra/tutorials/complex_numbers.html

thank you, I'll check it.
I hope it helps!

Not sure if its correct but i got
(9+6i)(15i-20)
where i= sqrt of -1

I meant (9+6i)/(15i-20)

Ok, I do finance not maths, so this can be ridden with errors. Also typed on my phone so ignore the capitalization of 'I'.

I/5 - ((2-2i)/4-3i)) = 0
I/5 = (2-2i)/(4-3i)
I/5(4-3i) = (2-2i)
I(4-3i)/5 = (2-2i)
I(4-3i) = 5 (2-2i)
4i - 3i^2 = 10 - 10i
4i - 3i^2 + 10i = 10
-3i^2 + 14i - 10 = 0
Using almighty formula (-b±*sq root of* b^2-4ac/2a)
-14 ± *sq root of* -14^2 - 4 (-3*-10)/2*-3
-14 ± *sq root of* 196 - 120/-6
-14 ± 8.72 / -6
-14 ± -1.45
I = -14 - 1.45 or I = -14 + 1.45
I = -15.45 or I = -12.55

I doubt this is correct, probably because the quadratic cannot be solved by factorization. But at least this is one attempt. Hopefully the correct answer can be found if you guys correct whatever errors you find.

Sammy107_d:

Ok, I do finance not maths, so this can be ridden with errors. Also typed on my phone so ignore the capitalization of 'I'.

I/5 - ((2-2i)/4-3i)) = 0
I/5 = (2-2i)/(4-3i)
I/5(4-3i) = (2-2i)
I(4-3i)/5 = (2-2i)
I(4-3i) = 5 (2-2i)
4i - 3i^2 = 10 - 10i
4i - 3i^2 + 10i = 10
-3i^2 + 14i - 10 = 0
Using almighty formula (-b±*sq root of* b^2-4ac/2a)
-14 ± *sq root of* -14^2 - 4 (-3*-10)/2*-3
-14 ± *sq root of* 196 - 120/-6
-14 ± 8.72 / -6
-14 ± -1.45
I = -14 - 1.45 or I = -14 + 1.45
I = -15.45 or I = -12.55

I doubt this is correct, probably because the quadratic cannot be solved by factorization. But at least this is one attempt. Hopefully the correct answer can be found if you guys correct whatever errors you find.

bros sorry but this is complex number not ordinary algebra. What u see as "i" is not actually 'i'. "I" is sqrt of -1. Since it doesnt have any mathematical value, mathematicians and engineers decided call it "i" or "j". So it is not a variable,its a constant.

Oh ok. That's interesting to learn. Not all letters are random like 'x'. Thanks

Thanks to you all for your replies,
I have been able to solve it.
The answer is
-14/25+7/25i