Author Topic: TrueTears question thread (Read 66220 times) Tweet Share

Damo17 · « **Reply #30 on:** December 27, 2008, 09:50:32 pm »

Quote from: TrueTears on December 27, 2008, 09:08:25 pm

no range was given it was just like that, so im guessing both answers are correct.

Also, this question:

If $z = 1 + i$ is a zero of the polynomial $z^3+az^2+bz+10-6i$ , find the values of $a$ and $b$ given that they are real.

Well this is a fairly easy question, since the coefficients are real, so 2 of the roots would be $1 + i$ and $1 - i$ . (conjugate pairs.)

So u sub both those in the former equation and solve the 2 simultaneous equations, however i keep getting $a = 3-3i$ and $b = -8+6i$ , even after solving on calc. But, the question says they are real numbers. Can anyone see what's wrong?

Sub in $z=1+i$ .

$z^3+az^2+bz+10-6i$

$= (1+i)^3+a(1+i)^2+b(1+i)+10-6i$

$=-2+2i+a2i+b+ib+10-6i$

$-2+b+10=0$

      $b=-8$

and $2i+a2i+ib-6i=0$
$2i+a2i-8i-6i=0$
         $a2i=14i$
      $a=6$

TrueTears · « **Reply #31 on:** December 27, 2008, 11:23:00 pm »

ah yeah, i just got that answer as well xD

and also, this question:

Find the solutions of the equation $z^4-2z^2+4=0$ in polar form.

just not sure how to factorise it...

humph · « **Reply #32 on:** December 27, 2008, 11:26:39 pm »

Make the substitution $w = z^2$ first, and solve the corresponding quadratic. Then substitute back in $z = \pm \sqrt{w}$ for your four solutions.

TrueTears · « **Reply #33 on:** December 27, 2008, 11:32:54 pm »

ahh yes thanks humph i got it.

TrueTears · « **Reply #34 on:** December 28, 2008, 01:06:37 am »

another 2 Q's

1. Prove that for any complex number $z$ , $3|z-1|^2 = |z+1|^2$ if and only if
$|z-2|^2=3$

Mao · « **Reply #35 on:** December 28, 2008, 01:51:14 am »

1. let z=x+yi

$3|z-1|^2 = |z+1|^2$

$\implies 3\times |(x-1)+yi|^2 = |(x+1)+yi|^2$

$\implies 3 \times \left( (x-1)^2 + y^2) = (x+1)^2 + y^2$

$\implies 3x^2 - 6x + 3 + 3y^2 = x^2 + 2x + 1 + y^2$

$\implies 2x^2 - 8x +2 + 2y^2 = 0$

$\implies x^2 - 4x +1 + y^2 = 0$

$\implies (x-2)^2 -3 + y^2 = 0$

$\implies |(x-2)+yi|^2 = 3$

$\implies |z-2|^2 = 3$

Mao · « **Reply #36 on:** December 28, 2008, 01:57:48 am »

TrueTears · « **Reply #37 on:** December 28, 2008, 02:01:09 am »

thanks mao for that proof, lol i dono why i posted that other Q up there, i got it straight after i posted LOL, probs too late in the night o.O

TrueTears · « **Reply #38 on:** December 28, 2008, 01:30:00 pm »

1. On the Argand plane sketch $|\frac{z-1-i}{z}|=1$

how do i simplify $|\frac{z-1-i}{z}|=1$ so i can sketch it?

2. If the real part of $\frac{z-1}{z+1}$ is zero, find the locus of points representing z in the complex plane.

Thanks

humph · « **Reply #39 on:** December 28, 2008, 01:51:45 pm »

Quote from: TrueTears on December 28, 2008, 01:30:00 pm

1. On the Argand plane sketch $|\frac{z-1-i}{z}|=1$

how do i simplify $|\frac{z-1-i}{z}|=1$ so i can sketch it?

As you know that $z \neq 0$ , you can multiply both sides by $|z|$ . So you have that
$|z-1-i| = |z|$
$\Longrightarrow (x-1)^2 + (y-1)^2 = x^2 + y^2$
$\Longrightarrow x^2 - 2x + 1 + y^2 - 2y + 1 = x^2 + y^2$
$\Longrightarrow y = 1 - x$

Quote from: TrueTears on December 28, 2008, 01:30:00 pm

2. If the real part of $\frac{z-1}{z+1}$ is zero, find the locus of points representing z in the complex plane.

Thanks

$\frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy}$
$\qquad = \frac{((x-1)+iy)((x+1)-iy)}{((x+1)+iy)((x+1)-iy)}$
$\qquad = \frac{x^2 - 1 + 2iy + y^2}{(x+1)^2 + y^2}$
$\qquad = \frac{x^2 - 1 + y^2}{(x+1)^2 + y^2} + i\frac{2y}{(x+1)^2 + y^2}$
Now the real part of this is zero, which is the first term. This means that the numerator of the first term is equal to zero; that is,
$x^2 - 1 + y^2 = 0$
$\Longrightarrow x^2 + y^2 = 1$
So the locus of points representing z for which the real part of $\frac{z-1}{z+1}$ is zero is merely the unit circle in the complex plane.

Interestingly enough, a question very similar to this was on my complex analysis exam last semester! (Mind you, it was the very first question, a "warm-up question" if you will.)

TrueTears · « **Reply #40 on:** December 28, 2008, 01:57:00 pm »

ahhh thanks humph, so wat happens to the $i\frac{2y}{(x+1)^2 + y^2}$ ?

humph · « **Reply #41 on:** December 28, 2008, 02:08:17 pm »

Oh yeah, I forgot. You can't have that $z=-1$ as otherwise $\frac{z-1}{z+1}$ is undefined. So the answer actually should be $\{z \in \mathbb{C} : |z|=1, \; z \neq -1 \}$ .

The $i\frac{2y}{(x+1)^2 + y^2}$ is just the imaginary part of $\frac{z-1}{z+1}$ ; it's unimportant in this question.

TrueTears · « **Reply #42 on:** December 28, 2008, 02:12:37 pm »

ahhh, thanks that clears it all up xD

TrueTears · « **Reply #43 on:** December 28, 2008, 05:42:31 pm »

and...

If $Arg z = \frac{\pi}{4}$ and $Arg(z-3) = \frac{\pi}{2}$ , find $Arg(z-6i})$ .

vce08 · « **Reply #44 on:** December 28, 2008, 06:03:23 pm »

Quote from: TrueTears on December 28, 2008, 05:42:31 pm

and...

If $Arg z = \frac{\pi}{4}$ and $Arg(z-3) = \frac{\pi}{2}$ , find $Arg(z-6i})$ .

Is the answer - pi/4?

Search the forums now!

We have moved!

Author Topic: TrueTears question thread (Read 66220 times) Tweet Share

Damo17

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

humph

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

Mao

Re: TrueTears question thread

Mao

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

humph

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

humph

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

vce08

Re: TrueTears question thread

Recent Posts

User:
Password: