# complex numbers p4

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#1
hey can anyone help me with this one question?
ok well here goes

gven that z=2-i and z^2 = 3-4i

hence or otherwise find the roots, z1 and z2 of the equation

(z+i)^2 = 3-4i

display these roots on an argand diagram
(a) deduce tht |z1-z2| = 2(root5)
b) find the value of arg(z1+z2)

help!its in the p4 book by heineman page 37 question 23! its a past exam question and i cant do it!! 0
15 years ago
#2
(Original post by kriztinae)
hey can anyone help me with this one question?
ok well here goes

gven that z=2-i and z^2 = 3-4i

hence or otherwise find the roots, z1 and z2 of the equation

(z+i)^2 = 3-4i

display these roots on an argand diagram
(a) deduce tht |z1-z2| = 2(root5)
b) find the value of arg(z1+z2)

help!its in the p4 book by heineman page 37 question 23! its a past exam question and i cant do it!! it seems wrong
(z+i)^2 = (2-i+i)^2 = 4 =/= 3-4i
0
15 years ago
#3
(Original post by keisiuho)
it seems wrong
(z+i)^2 = (2-i+i)^2 = 4 =/= 3-4i
how would you do it though?

i tried getting rid of the i's using various substituition then came to z = (2-+root(2))/2
0
#4
(Original post by keisiuho)
it seems wrong
(z+i)^2 = (2-i+i)^2 = 4 =/= 3-4i
thank you!! thats what ithought!
ill leave it! thanx
0
15 years ago
#5
I think it should be:
gven that if z=2-i then z^2 = 3-4i
and the z in the equation is not necessarily 2-i

so (z+i)^2 = 3-4i
(z+i)^2 = (2-i)^2
z+i = 2-i OR z+i = i-2
z = 2-2i OR -2
0
#6
(Original post by keisiuho)
I think it should be:
gven that if z=2-i then z^2 = 3-4i
and the z in the equation is not necessarily 2-i

so (z+i)^2 = 3-4i
(z+i)^2 = (2-i)^2
z+i = 2-i OR z+i = i-2
z = 2-2i OR -2
hmm maybe, ill keep trying
0
15 years ago
#7
(Original post by kriztinae)
hey can anyone help me with this one question?
ok well here goes

gven that z=2-i and z^2 = 3-4i

hence or otherwise find the roots, z1 and z2 of the equation

(z+i)^2 = 3-4i

display these roots on an argand diagram
(a) deduce tht |z1-z2| = 2(root5)
b) find the value of arg(z1+z2)

help!its in the p4 book by heineman page 37 question 23! its a past exam question and i cant do it!! (z+i)^2 = 3-4i
z² + 2zi + i² = 3 - 4i
z² + 2zi + 4i - 4 = 0
z² + z(2i + 4) - 4 = 0
(z + (i+2))² - (2i + 4)² - 4 = 0
(z + (i+2))² - (4i² + 16i + 16) - 4 = 0
(z + (i+2))² + 4 - 16i - 20 = 0
(z + (i+2))² = 16(1 + i)
z + (i+2) = ±4√(1 + i)
z = i+2 ±4√(1 + i)

z1 = i+2+4√(1 + i)
z2 = i+2-4√(1 + i)

|z1-z2| = |(i+2+4√(1 + i)) - (i+2-4√(1 + i))|
|z1-z2| = |(8√(1 + i)| =/= 2√5

lol, I don't think I'm ready for these P4 questions. 0
#8
(Original post by mik1a)
(z+i)^2 = 3-4i
z² + 2zi + i² = 3 - 4i
z² + 2zi + 4i - 4 = 0
z² + z(2i + 4) - 4 = 0
(z + (i+2))² - (2i + 4)² - 4 = 0
(z + (i+2))² - (4i² + 16i + 16) - 4 = 0
(z + (i+2))² + 4 - 16i - 20 = 0
(z + (i+2))² = 16(1 + i)
z + (i+2) = ±4√(1 + i)
z = i+2 ±4√(1 + i)

z1 = i+2+4√(1 + i)
z2 = i+2-4√(1 + i)

|z1-z2| = |(i+2+4√(1 + i)) - (i+2-4√(1 + i))|
|z1-z2| = |(8√(1 + i)| =/= 2√5

lol, I don't think I'm ready for these P4 questions. erm thanx!
0
15 years ago
#9
(z+i)^2 = 3 - 4i
z + i = + or - sqrt(3-4i)
It's given that sqrt(3-4i) = 2 - i
so z = -i + or - (2 - i)
so the 2 roots are z1 = 2(1-i) and z2 = -2
0
15 years ago
#10
yup take the root of both sides as you're given the z^2 = 3-4i and that z=2-i

so if you root both sides,

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

so

z+i = 2 - i
z+i = -2 + i

z = 2 - 2i or -2

which are easy to draw on argand diagram 0
15 years ago
#11
if youre given z = some sort of fraction, is there an easier way to find |z| and argz than having to rationalise ?
0
15 years ago
#12
(Original post by kikzen)
if youre given z = some sort of fraction, is there an easier way to find |z| and argz than having to rationalise ?
|z/w| = |z|/|w| and arg(z/w) = argz - argw
0
15 years ago
#13
(Original post by Bezza)
|z/w| = |z|/|w| and arg(z/w) = argz - argw
also true for multiplication
0
15 years ago
#14
(Original post by keisiuho)
also true for multiplication
Except with a * and + rather than / and - 0
15 years ago
#15
Is it also true for powers then?

eg.

does arg(|z|)^|w| = arg |z| * arg |w|?
0
15 years ago
#16
(Original post by mik1a)
Is it also true for powers then?

eg.

does arg(|z|)^|w| = arg |z| * arg |w|?
I'm not sure why you have all the moduli signs in there - |z| would be a real number so arg|z| would be 0 or pi.

|zw| = |z||w| and arg(zw) = argz + argw

arg(z^n) = arg(z*z*z*z*z*.....*z) = n*argz
0
15 years ago
#17
(Original post by Bezza)
I'm not sure why you have all the moduli signs in there - |z| would be a real number so arg|z| would be 0 or pi.

|zw| = |z||w| and arg(zw) = argz + argw

arg(z^n) = arg(z*z*z*z*z*.....*z) = n*argz
arg(z) = arg(z^(1/n)*z^(1/n)*z^(1/n)*z^(1/n)*...*z^(1/n)) = n*arg(z^(1/n))
so arg(z^(1/n)) = 1/n arg(z) 0
15 years ago
#18
so uh just like logs then eh
0
15 years ago
#19
(Original post by kikzen)
so uh just like logs then eh
That's what I was thinking when I asked that question... although now that I think about it,

(log a)^b =/= (log a)(log b)

Silly me.
0
15 years ago
#20
wot does arg mean?
0
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