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FP1 complex number

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?
If (z+i)^2= 3-4i then z+i =the roots from part a


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(edited 8 years ago)
Reply 2
Avatar for PLM98
PLM98
6
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!
Reply 3
Avatar for TeeEm
TeeEm
19
you can do part (a) and you cannot do part (b)? ....
Reply 4
Avatar for alesha98
alesha98
OP
4
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?
Reply 5
Avatar for TeeEm
TeeEm
19
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
Reply 6
Avatar for alesha98
alesha98
OP
4
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
Reply 7
Avatar for alesha98
alesha98
OP
4
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
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
Reply 9
Avatar for alesha98
alesha98
OP
4
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 :smile:
Reply 10
Avatar for TeeEm
TeeEm
19
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)
Reply 11
Avatar for alesha98
alesha98
OP
4
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
Reply 12
Avatar for TeeEm
TeeEm
19
Original post by alesha98
ok thank you so much


let me scan an example I made earlier ...
different approaches
Reply 13
Avatar for TeeEm
TeeEm
19
Original post by alesha98
ok thank you so much


..
IMG_0001.jpg398.6KB
IMG_0002.jpg339.1KB
Reply 14
Avatar for -Bex-
-Bex-
10
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 :tongue:

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