Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

Ricky
2006-01-09 10:14:47

The reason why I did that is because x and x*, and z and z*, have a lot of the same terms in them, which you can take advantage of if you put them in full complex form.

From there, it's just plugging them in and trying to solve for c and d.

Peter T
2006-01-09 07:29:54

thanks, i sort of understood what u did. ill be going along to my math support class to understand that fully.

Ricky
2006-01-09 06:45:34

First, start by doing a few quick substitutions:

x   = a + bi (where a and b are real)
x* = a - bi

z   = c + di (where c and d are real)
z* = c - di

xx* = (a + bi)(a - bi) = a^2 + b^2
z - z* = c + di - (c - di) = 2di

So:

a^2 + b^2 + 3(2di) = 13 + 12i

Since i is the only possible imaginary value (all other variables must be real):

a^2 + b^2 = 13 and 6di = 12i

di = 2i, and since d must be real, d = 2

So z = c + 2i, where c is any real number.  The simpilest solution is z = 2i (c = 0)

Peter T
2006-01-09 04:56:39

Question:
Find z such that
xx* + 3(z-z*) = 13 + 12i

can someone please tell me how to begin to solve this.

i realize that z* is the complex conjigate of z, but i just don't know where to start.

Thanks

Board footer

Powered by FluxBB