# complex numbers p4

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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!!

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!!

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#2

(Original post by

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!!

**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 = (2-i+i)^2 = 4 =/= 3-4i

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#3

(Original post by

it seems wrong

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

**keisiuho**)it seems wrong

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

i tried getting rid of the i's using various substituition then came to z = (2-+root(2))/2

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(Original post by

it seems wrong

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

**keisiuho**)it seems wrong

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

ill leave it! thanx

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#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

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

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(Original post by

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

**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

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#7

**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!!

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.

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(Original post by

(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.

**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.

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#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

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

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#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

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

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#11

if youre given z = some sort of fraction, is there an easier way to find |z| and argz than having to rationalise ?

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#12

(Original post by

if youre given z = some sort of fraction, is there an easier way to find |z| and argz than having to rationalise ?

**kikzen**)if youre given z = some sort of fraction, is there an easier way to find |z| and argz than having to rationalise ?

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#13

(Original post by

|z/w| = |z|/|w| and arg(z/w) = argz - argw

**Bezza**)|z/w| = |z|/|w| and arg(z/w) = argz - argw

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#14

(Original post by

also true for multiplication

**keisiuho**)also true for multiplication

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#16

(Original post by

Is it also true for powers then?

eg.

does arg(|z|)^|w| = arg |z| * arg |w|?

**mik1a**)Is it also true for powers then?

eg.

does arg(|z|)^|w| = arg |z| * arg |w|?

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

arg(z^n) = arg(z*z*z*z*z*.....*z) = n*argz

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#17

(Original post by

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

**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

so arg(z^(1/n)) = 1/n arg(z)

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#19

(Original post by

so uh just like logs then eh

**kikzen**)so uh just like logs then eh

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

Silly me.

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