Quick Complex Numbers Q (1 Viewer)

nrlwinner · Dec 13, 2009

z1 and z2 are two complex numbers whose argument differ by 90 degrees. Prove mod (z1 + z2) = mod (z1 - z2)

I know this has something to do with complex conjugates, but I dunno how to apply it,

kaz1 · Dec 13, 2009

Use an Argrand Diagram with a rectangle, a property of a rectangle is that the diagonals are equal which are z₁ + z₂ and z₁-z₂

study-freak · Dec 13, 2009

Lukybear · Dec 13, 2009

study-freak said:
$\\$Let $z_1=kiz_2$ (Arguments differ by 90^\circ ) \\\left | z_1+z_2 \right | \\=\left | z_2(1+ki) \right | \\=\left | z_2 \right |\left | 1+ki \right | \\=\left | z_2 \right |\sqrt{1+k^2} \\=\left | z_2 \right |\left | ki-1 \right | \\=\left | kiz_2-z_2 \right | \\=\left | z_1-z_2 \right |$

Would that method be completely right? Because if z1=kiz2, then the length would be changed. i.e. if z1=2iz2, then mod z1 =/ mod z2.

Wouldnt it be right algebracally to take both cases, of positive and negative?

study-freak · Dec 13, 2009

Lukybear said:
Would that method be completely right? Because if z1=kiz2, then the length would be changed. i.e. if z1=2iz2, then mod z1 =/ mod z2.

Wouldnt it be right algebracally to take both cases, of positive and negative?

Yea, but we are not dealing with mod z1 or mod z2
but mod(z1+z2) and mod(z1-z2)

And what do you mean by positive and negative cases?

Lukybear · Dec 13, 2009

oo, nvm. Thxs for that. I was thinking graphically about a //gram.

And i must admit, that algebric way is very elegant. Very nicely done study-freak.

study-freak · Dec 13, 2009

Lukybear said:
oo, nvm. Thxs for that. I was thinking graphically about a //gram.

And i must admit, that algebric way is very elegant. Very nicely done study-freak.

Haha, thanks for that. I intuitively tried this way.

nrlwinner · Dec 13, 2009

study-freak said:
$\\$Let $z_1=kiz_2$ (Arguments differ by 90^\circ ) \\\left | z_1+z_2 \right | \\=\left | z_2(1+ki) \right | \\=\left | z_2 \right |\left | 1+ki \right | \\=\left | z_2 \right |\sqrt{1+k^2} \\=\left | z_2 \right |\left | ki-1 \right | \\=\left | kiz_2-z_2 \right | \\=\left | z_1-z_2 \right |$

Thanks for the help, but I'm lost after the root.

study-freak · Dec 13, 2009

nrlwinner · Dec 13, 2009

Thanks for the help. I've been able to finish all my homework now, except this.

How would you prove algebraically that z2/z1 is purely imagenary?

study-freak · Dec 13, 2009

nrlwinner · Dec 13, 2009

study-freak said:
$\\$Since $z_1=kiz_2 \\\frac{z_2}{z_1}=\frac{1}{ki} \\=\frac{-i}{k}$ which is purely imaginary$$

Oops, forgot to state that the 90 degrees is not given. I'm doing this backwards.

study-freak · Dec 13, 2009

nrlwinner said:
Oops, forgot to state that the 90 degrees is not given. I'm doing this backwards.

lol, um, post the full question?

nrlwinner · Dec 13, 2009

study-freak said:
lol, um, post the full question?

mod (z1 - z2) = mod (z1 + z2)

Prove z2/z1 is purely imagenary.

study-freak · Dec 13, 2009

nrlwinner · Dec 13, 2009

Thanks freak. Can someone help me out here. I don't know how to finish this off.

$\frac{1-cos\Theta +isin\Theta }{1+cos\Theta +isin\Theta }$

$\frac{2sin^2\frac{\Theta }{2}+i2sin\frac{\Theta }{2}cos\frac{\Theta }{2}}{2cos^2\frac{\Theta }{2}+i2sin\frac{\Theta }{2}cos\frac{\Theta }{2}}$

$\frac{2sin\frac{\Theta }{2}(sin\frac{\Theta }{2}+icos\frac{\Theta }{2})}{2cos\frac{\Theta }{2}(cos\frac{\Theta }{2}+isin\frac{\Theta }{2})}$

$tan\frac{\Theta }{2}.\frac{sin\frac{\Theta }{2}+icos\frac{\Theta }{2}}{cos\frac{\Theta }{2}+isin\frac{\Theta }{2}}$

Im trying to prove for $-itan\frac{\Theta }{2}$

gurmies · Dec 13, 2009

cyl123 · Dec 13, 2009

I think you got the question wrong. It should be (1-cosx-isinx)/(1+cosx+isinx) (test x=pi/2 to see the original is incorrect)
So you should get:
tan(x/2)(sin(x/2)-icos(x/2))/(cos(x/2)+isin(x/2))
=tan(x/2)(cos(pi/2-x/2)-isin(pi/2-x/2))/cis(x/2)
noting cos(x)=cos(-x) and -sin(x)=sin(-x)
=tan(x/2)(cos(-pi/2+x/2)+isin(-pi/2+x/2)/cis(x/2)
=tan(x/2)(cis(-pi/2+x/2))/cis(x/2)
noting cis(x)/cis(y)=cis(x-y)
=tan(x/2)cis(-pi/2+x/2-x/2)
=tan(x/2)cis(-pi/2) noting cis(-pi/2)=-i gives the result

EDIT: put it on latex

NewiJapper · Dec 13, 2009

i have absolutely NO IDEA what you guys are on about. It just looks like a bunch of Z's and I's haha.

I rule at intergration though. Possibly the most fun i have ever had in maths.

Lukybear · Dec 13, 2009

nrlwinner said:
Thanks freak. Can someone help me out here. I don't know how to finish this off.

$\frac{1-cos\Theta +isin\Theta }{1+cos\Theta +isin\Theta }$

$\frac{2sin^2\frac{\Theta }{2}+i2sin\frac{\Theta }{2}cos\frac{\Theta }{2}}{2cos^2\frac{\Theta }{2}+i2sin\frac{\Theta }{2}cos\frac{\Theta }{2}}$

$\frac{2sin\frac{\Theta }{2}(sin\frac{\Theta }{2}+icos\frac{\Theta }{2})}{2cos\frac{\Theta }{2}(cos\frac{\Theta }{2}+isin\frac{\Theta }{2})}$

$tan\frac{\Theta }{2}.\frac{sin\frac{\Theta }{2}+icos\frac{\Theta }{2}}{cos\frac{\Theta }{2}+isin\frac{\Theta }{2}}$

Im trying to prove for $-itan\frac{\Theta }{2}$

That require 3unit trig functions rite? If we dont have it we cannot do it?

Quick Complex Numbers Q (1 Viewer)

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