4u complex (1 Viewer)

jnney · Oct 16, 2011

1. if z = x+iy, find a relation between x and y given that re (z-4)/z = 0

2. find the complex number z such that z/(1+z) = 1 + i

3. if z = 1+i is a root of x^2+ax+b=0 find the real numbers a and b

4. given Re(z)= 3Imz, find z if z^2-6i is real.

nightweaver066 · Oct 16, 2011

jnney said:
1. if z = x+iy, find a relation between x and y given that re (z-4)/z = 0

2. find the complex number z such that z/(1+z) = 1 + i

3. if z = 1+i is a root of x^2+ax+b=0 find the real numbers a and b

4. given Re(z)= 3Imz, find z if z^2-6i is real.

1. $Re(\frac{z-4}{z}) = Re(\frac{x + iy - 4}{x + iy})$

$= Re(\frac{(x + iy - 4)(x-iy)}{(x + iy)(x-iy)}))$

$= Re(\frac{x^2 -4x + y^2 + i(x -xy +4)}{x^2 +y^2})$

$=\frac{x^2 -4x + y^2}{x^2 +y^2} = 0$

$\therefore x^2 + y^2 - 4x = 0$

2. $\frac{z}{1+z} = 1 + i$

$z = (1 +i)(1+z)$

$z = 1 + z + i + iz$

$iz = -1 - i$

$z = \frac{-1-i}{i}$

$z = \frac{-(1+i)\times -i}{i\times -i}$

$z = -1 + i$

3. $x^2 + ax + b = 0$

$\text{Sub x = 1 + i}$

$(1+i)^2 + a(1+i) + b =0$

$2i + a + ai + b =0$

$a + b + i(a + 2) = 0$

$\text{Equating real and unreal coefficients,}$

$a + b = 0....(1)$

$a + 2 = 0....(2)$

$\therefore a = -2$

$\text{From (1), } -2 + b = 0$

$\therefore b = 2$

4. $Re(z) = 3Im(z)$

$\therefore x = 3y$

$z = x + iy$

$z^2 - 6i = (x+iy)^2 - 6i$

$= x^2 - y^2 + 2ixy - 6i$

$= x^2 - y^2 + i(2xy - 6)$

$\text{For } z^2 - 6i\text{ to be real, } 2xy - 6 = 0$

$xy = 3$

$\therefore 3y^2 = 3$

$y = \pm1$

$\therefore x = \pm3$

$z = 3 + i\text{ or } z = -3 -i$

Deep Blue · Oct 16, 2011

For the first part, finish the question by completing the square and then describing the relationship as a circle with centre...

nightweaver066 · Oct 16, 2011

Deep Blue said:
For the first part, finish the question by completing the square and then describing the relationship as a circle with centre...

I don't think there's a need to. The question just asked to find a relationship, not describe the geometric behaviour of the relationship.

gurmies · Oct 16, 2011

nightweaver066 said:
1. $Re(\frac{z-4}{z}) = Re(\frac{x + iy - 4}{x + iy})$

$= Re(\frac{(x + iy - 4)(x-iy)}{(x + iy)(x-iy)}))$

$= Re(\frac{x^2 -4x + y^2 + i(x -xy +4)}{x^2 +y^2})$

$=\frac{x^2 -4x + y^2}{x^2 +y^2} = 0$

$\therefore x^2 + y^2 - 4x = 0$

2. $\frac{z}{1+z} = 1 + i$

$z = (1 +i)(1+z)$

$z = 1 + z + i + iz$

$iz = -1 - i$

$z = \frac{-1-i}{i}$

$z = \frac{-(1+i)\times -i}{i\times -i}$

$z = -1 + i$

3. $x^2 + ax + b = 0$

$\text{Sub x = 1 + i}$

$(1+i)^2 + a(1+i) + b =0$

$2i + a + ai + b =0$

$a + b + i(a + 2) = 0$

$\text{Equating real and unreal coefficients,}$

$a + b = 0....(1)$

$a + 2 = 0....(2)$

$\therefore a = -2$

$\text{From (1), } -2 + b = 0$

$\therefore b = 2$

4. $Re(z) = 3Im(z)$

$\therefore x = 3y$

$z = x + iy$

$z^2 - 6i = (x+iy)^2 - 6i$

$= x^2 - y^2 + 2ixy - 6i$

$= x^2 - y^2 + i(2xy - 6)$

$\text{For } z^2 - 6i\text{ to be real, } 2xy - 6 = 0$

$xy = 3$

$\therefore 3y^2 = 3$

$y = \pm1$

$\therefore x = \pm3$

$z = 3 + i\text{ or } z = -3 -i$

For question 3, it's a lot simpler to use the fact that a and b are real. Then, it must be true that the roots are 1+i and 1-i (complex conjugates). Sum of roots = -a = 1+i+1-i = 2 <===> a = -2 and product of roots = b = (1-i)(1+i) = 2.

nightweaver066 · Oct 16, 2011

gurmies said:
For question 3, it's a lot simpler to use the fact that a and b are real. Then, it must be true that the roots are 1+i and 1-i (complex conjugates). Sum of roots = -a = 1+i+1-i = 2 <===> a = -2 and product of roots = b = (1-i)(1+i) = 2.

Thanks for that. I've only started complex numbers myself and so i have not yet learnt the conjugate root theorem but i guess it's never too early to start applying it.

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