Imaginay Nos (1 Viewer)

shaon0 · Jan 25, 2010

jm01 said:
Or you could use long division...

His methods a lot better than long division and more efficient when one gets used to the equating co-effs poly method.

cutemouse · Jan 25, 2010

shaon0 said:
His methods a lot better than long division and more efficient when one gets used to the equating co-effs poly method.

"Better" is a subjective term.

Personally, I'd prefer to use long division. But I was just pointing out that there is an alternative method.

Lukybear · Jan 25, 2010

How is long divison achieved? Could you show via a eaxmple?

Lukybear · Jan 25, 2010

Also this question:
$Z=\frac{2+z}{2-z}$
. Show that as the point z describes the y axis, from the negative end to the positive end, the point Z (upper case) describes completely the circle x^2+y^2=1, in the coutner-clockwise sense.

Ive got x^2+y^2=4 as the circle, and not the question stated x^2+y^2=1. Can anyone confirm?

Lukybear · Jan 25, 2010

Also this question.

Prove that if z lies on the circle x^2+y^2=1, the points representing
$Z=\sqrt(\frac{1+z}{1-z})$
lies on an orthogonal line pair.

Lukybear · Jan 25, 2010

One more:

If P, Q represent the complex no.s z, Z and
$Z=\frac{1}{z-3} + \frac{17}{3}$

find the locus of Q as P moves on the circle |z-3|=3

cutemouse · Jan 25, 2010

Lukybear said:
How is long divison achieved? Could you show via a eaxmple?

Look in the Coroneos 4U book. It's there in one of the examples.

untouchablecuz · Jan 25, 2010

Lukybear said:
Also this question:
$Z=\frac{2+z}{2-z}$
. Show that as the point z describes the y axis, from the negative end to the positive end, the point Z (upper case) describes completely the circle x^2+y^2=1, in the coutner-clockwise sense.

Ive got x^2+y^2=4 as the circle, and not the question stated x^2+y^2=1. Can anyone confirm?

$\\Z=\frac{2+z}{2-z}\\z=2 \times \frac{Z-1}{Z+1}\\$Let $Z=x+iy,\\z=2 \times \frac{x+iy-1}{x+iy+1}=2 \times \frac{(x^2+y^2-1)+i(2y)}{(x+1)^2+y^2}\\$If $z $ describes the $y$-axis, then $ \Re (z)=0\\\therefore x^2+y^2-1=0\Rightarrow x^2+y^2=1\\$Hence the point $Z $ describes a circle with radius $1 $ and origin at $0+0i$

untouchablecuz · Jan 25, 2010

Lukybear said:
Also this question.

Prove that if z lies on the circle x^2+y^2=1, the points representing
$Z=\sqrt(\frac{1+z}{1-z})$
lies on an orthogonal line pair.

$\\Z=\sqrt\frac{1+z}{1-z}\Rightarrow z=\frac{Z^2-1}{Z^2+1}\\$If $z $ lies on the circle $x^2+y^2=1, $ then, $|z|=1\\\therefore |z|=\left|\frac{Z^2-1}{Z^2+1}\right|=1\Rightarrow |Z^2-1|=|Z^2+1|\\$

not 100% sure how to proceed

khorne · Jan 25, 2010

Where are you even getting these questions from? They don't look very standard to me.

untouchablecuz · Jan 25, 2010

Lukybear said:
One more:

If P, Q represent the complex no.s z, Z and
$Z=\frac{1}{z-3} + \frac{17}{3}$

find the locus of Q as P moves on the circle |z-3|=3

$\\P: \ z=a+ib \ (*)\\Q: \ \frac{1}{z-3}+\frac{17}{3}=X+iY=Z \ (**)\\$From $ (**),\\X=\frac{a-3}{(a-3)^2+b^2}+\frac{17}{3} $ and $ Y=-\frac{b}{(a-3)^2+b^2}\\$But $ |z-3|=3\Rightarrow (a-3)^2+b^2=9 \ (***)\\\therefore X=\frac{a-3}{9}+\frac{17}{3} $ and $ Y=-\frac{b}{9}\\$Rearranging and squaring, $\\(a-3)^2=(9X-51)^2 $ and $b^2=(-9Y)^2=(9Y)^2\\$Sub in $(***),\\(9X-51)^2+(9Y)^2=9\\(X-\frac{51}{9})^2+Y^2=\frac{1}{9} \\\therefore |Z-\frac{51}{9}|=\frac{1}{3}$ is the locus of $ Q $ as $P $ moves on the circle $|z-3|=3$

not sure if this correct

Trebla · Jan 25, 2010

Lukybear said:
Also this question.

Prove that if z lies on the circle x^2+y^2=1, the points representing
$Z=\sqrt(\frac{1+z}{1-z})$
lies on an orthogonal line pair.

LOL, that is a term commonly used for vectors in a number space and matrix algebra...don't think it's that relevant to HSC

Lukybear · Jan 25, 2010

jm01 said:
Look in the Coroneos 4U book. It's there in one of the examples.

Which chapter? Its not in Imaginary thats a for sure.

Lukybear · Jan 25, 2010

Trebla said:
LOL, that is a term commonly used for vectors in a number space and matrix algebra...don't think it's that relevant to HSC

reali? I thought it was just y=+-x

gurmies · Jan 25, 2010

Orthogonal means at 90 degrees to.

$Z=\sqrt{\frac{1+z}{1-z}}, \,\, \text{with} \,\, \left |z \right |=1 \\\\ \text{Let} \,\, z=\cos2\theta+i\sin2\theta \\\\ Z=\sqrt{\frac{1+\cos2\theta+i\sin2\theta}{1-\cos2\theta-i\sin2\theta}} \\\\ =\sqrt{\frac{2\cos^{2}\theta+2i\sin\theta\cos\theta}{2\sin^{2}\theta-2i\sin\theta\cos\theta}} \\\\ = \sqrt{i\cot2\theta} \\\\ = \pm \sqrt{\cot2\theta}\text{cis}\frac{\pi}{4}$

Lukybear · Jan 25, 2010

untouchablecuz said:
$\\P: \ z=a+ib \ (*)\\Q: \ \frac{1}{z-3}+\frac{17}{3}=X+iY=Z \ (**)\\$From $ (**),\\X=\frac{a-3}{(a-3)^2+b^2}+\frac{17}{3} $ and $ Y=-\frac{b}{(a-3)^2+b^2}\\$But $ |z-3|=3\Rightarrow (a-3)^2+b^2=9 \ (***)\\\therefore X=\frac{a-3}{9}+\frac{17}{3} $ and $ Y=-\frac{b}{9}\\$Rearranging and squaring, $\\(a-3)^2=(9X-51)^2 $ and $b^2=(-9Y)^2=(9Y)^2\\$Sub in $(***),\\(9X-51)^2+(9Y)^2=9\\(X-\frac{51}{9})^2+Y^2=\frac{1}{9} \\\therefore |Z-\frac{51}{9}|=\frac{1}{3}$ is the locus of $ Q $ as $P $ moves on the circle $|z-3|=3$

not sure if this correct

Absolutely rite. Can i just ask, i did it just graphically. I was like: the locus of that was

The locus of 1/z-3 was a circle of same centre but with radius 1/3. Hence 1/z-3 + 17/3 equals to a locaus of (x-3-17/3)^2 + y^2 = 1/9

Where did i go wrong with that method?

Lukybear · Jan 25, 2010

gurmies said:
Orthogonal means at 90 degrees to.

Yep and y=+-x makes 90 degrees at centre.

khorne · Jan 25, 2010

Lukybear said:
Absolutely rite. Can i just ask, i did it just graphically. I was like: the locus of that was

The locus of 1/z-3 was a circle of same centre but with radius 1/3. Hence 1/z-3 + 17/3 equals to a locaus of (x-3-17/3)^2 + y^2 = 1/9

Where did i go wrong with that method?

Wouldn't the radius be the same, but the center be shifted, as 17/3 only shifts it?

Lukybear · Jan 25, 2010

khorne said:
Wouldn't the radius be the same, but the center be shifted, as 17/3 only shifts it?

So how would one determine the shifted centre?

khorne · Jan 25, 2010

Lukybear said:
So how would one determine the shifted centre?

On additional thought, I think you are right as to say that the radius changes too...

Which book are these problems from.

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