Complex numbers?

Question

Find all z which satisfy the following, expressing answers in x+iy form :

(i) (2+i)z + 3i - 1= 2 + 2i

(ii) (z-i)/(z-1) = 2/3

Any help will be appreciated

Answer 1

Qu. (i)
2z + iz = 3 - i
(2 + i) z = 3 - i
z = (3 - i) / (2 + i)
z = (3 - i)(2 - i) / 5
z = (6 - 5i + i²) / 5
z = (5 - 5i) / 5
z = 1 - i

Qu(ii)
3(z - i) = 2(z - 1)
3z - 3i = 2z - 2
z = - 2 + 3i

Answer 2

(i) (2+i)z +3i -1 = 2 + 2i
(2+i)z = 3 -i
z = (3 - i) / (2 + i) multiply top and bottom by (2 - i)
z = (6 - 5i + i^2) / (4 - i ^2) remember i^2 = -1
z = (6 - 5i -1) / (4 + 1)
z = (5 - 5i) / 5
z = 1 - i

(ii) (z - i) / (z - 1) = 2/3
3z - 3i = 2z -2
z = -2 + 3i

Answer 3

The sq. root of an imaginary quantity is yet another imaginary quantity. First notice that (a+bi)^2 = a^2 +2abi -b^2 = a^2-b^2 + (2ab)i So as an occasion, (2+i)^2 = 4 +4i -a million) =3+4i consequently the sqrt of three+4i is two+i and a pair of-a million because of the fact complicated numbers constantly are available conjugate pairs. So enable's see a thank you to tutor that the sqrt of three+4i is definitely 2 + i and a pair of - i keep in mind (a+bi)^2= a^2 -b^2 + (2ab)i So we could desire to tutor that a^2 - b^2 = 3 and 2ab = 4 placed b= 2/.a into first equation getting a^2-4/a^4=3 a^4 -3a^2 -4 =0This is a quadratic in a^2 which may be remedy by factoring giving a^2 = 4 or -a million. Reject the -a million as we'd like a genuine answer so x^2 = 4 and a = 2 or -2 b=2/a so b= a million or -a million for this reason the sqrt of three+=4i is two+i or 2-i'm hoping this helped.

Answer 4

1)(2+i)z=2+2i-3i+1
z=(3-i)/(2+i)
Multiply the LHS by the conjugate 2-i on both top and bottom
z=(6-5i-1)/(2+1)
z=(5-5i)/5
z=1-i

2)3(z-i)=2(z-1)
3z-3i=2z-2
z=-2+3i