/sci/ - Science & Math

expand the brackets of the first.

What does the z with a bar mean?

Those are linear equations. If you are older than 11 and can't solve them yourself, than you should just kill yourself.

z-bar is probably supposed to be the complex conjugate

>>3853889
Complex conjugate. <span class="math">\bar{a+ib}=a-ib[/spoiler] for a,b reals.

There must be a better way to do the 2nd one, but I'd start by writing everything in terms of a+ib... Someone will come up with a better way to do it for sure.

for the second one, replace z = a + bi to make it geyser.

>>3853895

I see, I've seen it usually written as z'.
In which case I would swap z for x+iy and z' for x-iy and when solve the real and imaginary for x and y

>>3853873
How about you do your own fucking homework?

>>3853893
>complex conjugation
>linear

Pick one.

>>3853902
what is the complex conducate of +?

>>3853911
Yeah, sorry, "\bar" didn't do what I expected.

New try:
<span class="math">\xbar{a+ib}[/spoiler]

I'll have to google if it doesn't work...

>>3853916
And it's \overline:
<span class="math">\overline{a+ib}[/spoiler]

Thank you, wikipedia copy/paste. And sorry for the mess.

for the second one, 2y+3x=12 and 3y-2x=-13, where z=x+yi

also, >>3853909
wut?

ty guys (;

>>3853941
good luck on your exam for this, you dumb whore

>>3853939
<span class="math">\overline{z_1z_2}=\overline{z_1}\overline{z_2}[/spoiler], not <span class="math">\overline{z_1z_2}=z_1\overline{z_2}[/spoiler] or <span class="math">\overline{z_1z_2}=\overline{z_1}z_2[/spoiler]. Complex conjugation is linear only when considering <span class="math">\mathbb{C}[/spoiler] as an <span class="math">\mathbb{R}[/spoiler] vector space. If you consider it as a <span class="math">\mathbb{C}[/spoiler] vector space, then it's not linear anymore. And when you talk about solving linear equations in <span class="math">\mathbb{C}[/spoiler], you certainly mean it as a <span class="math">\mathbb{C}[/spoiler] vector space.

>>3853950
There should have been a little space between the <span class="math">\overline{z_1}[/spoiler] and the <span class="math">\overline{z_2}[/spoiler] in my first equation.

>>3853952
but in OP's question, the conjugate is not multiplied, if you take z=x+yi you have 2i(x+yi)-3(x-yi)=-12+13i

multiply out the brackets, compare real parts and imaginary parts to form two linear equations. yes/no?

>>3853991
Yes I agree that you end up with a linear system by splitting <span class="math">z[/spoiler] into its real and imaginary parts. What I'm wondering is if the equation in <span class="math">z[/spoiler] can be called "linear", strictly speaking. I'd say it can't and it's not linear in <span class="math">z[/spoiler].

>>3854013
ahhhh now i see what you meant, i misinterpreted what you said, because i assumed that >>3853893
meant linear equations when comparing the real and imaginary parts after expansion

lot of derp in this thread :)

>>3854023
Yeah, maybe it's what he meant, indeed.