write the complex number 2-i/5+i in the usual a + bi form?

Answer 1

2-i/5+i = (2-i)*(5+i)ֿ¹

As (x+iy)ֿ¹ = (x+iy)*/||(x+iy)||²

(5+i)ֿ¹ = (5-i)/(5²+1²) = (5-i)/26

So (2-i)*(5+i)ֿ¹ = (2-i)(5-i)/26

= (10 -2i - 5i - 1)/26 = (9 -7i)/26 = 9/26 - (7/26)i

Another way to work it without having to know what the recipical is etc is this:

2-i/5+i = (x+iy) - as the answer can always be expressed an complex number

So

2 - i = (5+i)(x+iy) = 5x + 5yi + xi - y = 5x - y +(x+5y)i

Therefore this splits into two simultaneous equations:

5x - 1y = 2 ..............(1)
1x + 5y = -1 ............. (2)

Multiply (1) by 5 and add (1) and (2)

25x - 5y = 10 ..............(1)*5
1x + 5y = -1 ............. (2)
_________________________
26x + 0y = 9

So x = 9/26

Substitute this back into (1) to get y

5(9/26) - y = 2

y = 45/26 - 2 = (45-52)/26 = -7/26

So 2-i/5+i = 9/26 - (7/26)y

Answer 2

You need to multiply the numerator and denominator by the complex conjugate of the denominator. In this case, the denominator is 5 + i, so the conjugate is 5 - i. That gives you [(2 - i)(5 - i)]/[(5 + i)(5 - i)] = (10 - 5i - 2i + i^2)/(25 + 5i - 5i - i^2) = (10 - 7i - 1)/(25 + 1) = (9 - 7i)/26 = 9/26 - (7/26)i. This works because multiplying the denominator by its complex conjugate always gives a real number as the denominator.

Answer 3

2-i/5+i =
2 + (4/5)i

a=2
b=4/5
or did you mean (2-i)/(5+i)?
then multiply numerator and denominator by
(5-i) which in the denominator's complex conjugate. We have

[(2-i)(5-i)]/[(5+i)(5-i)]=
=(9 - 7i)/26
so
a= 9/26
b=-7/26

Answer 4

Conjugate of 5+i = 5 - i

Mutiply numerator and denominator of (2-i)/(5+i) by 5-i

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

= (10-7i-1)/25+1

= (9-7i)/26

=(9/26) - (7/26)i

Answer 5

(9/26) - (7/26)i

Answer 6

=[2+i]/[5+i]
=[2+i][5-i]/26
=[10+3i+1]/26
=11/26+[3/26]i
=.423+.115i