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# FP1 complex number watch

1. a) Given that z = 2 - i, show that z^2 = 3 - 4i.
b) Hence, or otherwise, find the roots, z1 and z2, of the equation (z+i)^2 = 3 - 4i.
I got part a right, but can someone teach me how i should start on part b please?
2. If (z+i)^2= 3-4i then z+i =the roots from part a

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3. From a we know z+i = sqrt(3-4i) so z+i=+ or - sqrt(3-4i) and hence z+i=+ or -(2-i) which gives you z=2-2i and z=-2

Hope it helps!
4. you can do part (a) and you cannot do part (b)? ....
5. (Original post by TeeEm)
you can do part (a) and you cannot do part (b)? ....
yes.
for part b, i started with (z-i)^2 = 3 - 4i
2-i = + or - sqrt(3-4i)
What should i do next? as i dont really get the question. When it said find the roots, does it mean find the square root of 3-4i?
6. (Original post by alesha98)
yes.
for part b, i started with (z-i)^2 = 3 - 4i
2-i = + or - sqrt(3-4i)
What should i do next? as i dont really get the question. When it said find the roots, does it mean find the square root of 3-4i?
teaching till 23.15

the others above are giving you the correct method
7. (Original post by TeeEm)
teaching till 23.15

the others above are giving you the correct method
sorry i am learning it myself and i just dont get this type of question, when it asks for the roots. in the case of part b, how does it show 2 - i is the root?
i am happy to stay till 23,15 to learn FP1, thankyou
8. (Original post by TeeEm)
teaching till 23.15

the others above are giving you the correct method
hi can someone check my working? i managed to work out the solution

Since from part a, i can get
z^2 = 3 - 4i
z = +or- sqrt (3-4i)
2-i = +or- sqrt (3-4i)
sqrt(3-4i) = 2-i or -2+i

so in part b,
(z+i)^2 = 3 - 4i
z+i = +or- sqrt (3-4i)
z+i = 2-i or -2+i
z = 2-2i or -2
9. (Original post by alesha98)
hi can someone check my working? i managed to work out the solution

Since from part a, i can get
z^2 = 3 - 4i
z = +or- sqrt (3-4i)
2-i = +or- sqrt (3-4i)
sqrt(3-4i) = 2-i or -2+i

so in part b,
(z+i)^2 = 3 - 4i
z+i = +or- sqrt (3-4i)
z+i = 2-i or -2+i
z = 2-2i or -2
if z = 2-i
z^2 = (2-i)^2

Second part is easy, put some thought into it
10. (Original post by LelouchViRuge)
if z = 2-i
z^2 = (2-i)^2

Second part is easy, put some thought into it
(2-i)^2 = (2-i)(2-i), i got this right at the begining
11. (Original post by alesha98)
sorry i am learning it myself and i just dont get this type of question, when it asks for the roots. in the case of part b, how does it show 2 - i is the root?
i am happy to stay till 23,15 to learn FP1, thankyou
I will write a similar question with a full solution to enlighten you, tomorrow (day off)
12. (Original post by TeeEm)
I will write a similar question with a full solution to enlighten you, tomorrow (day off)
ok thank you so much
13. (Original post by alesha98)
ok thank you so much
let me scan an example I made earlier ...
different approaches
14. (Original post by alesha98)
ok thank you so much
..
Attached Images

15. (Original post by alesha98)
i dont really get the question. When it said find the roots, does it mean find the square root of 3-4i?
I think what you were confused about is what a root is? Root just means solution,so it's asking for z, in case you hadn't worked that out yet

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