# in the set of complex #

• Oct 7th 2006, 06:07 AM
iceman1
in the set of complex #
A- 1) Solve the equation Z^2 – 5( 1+ i)z + 2+14i = 0

plz help me??
• Oct 7th 2006, 09:04 AM
AfterShock
Quote:

Originally Posted by iceman1
A- 1) Solve the equation Z^2 – 5( 1+ i)z + 2+14i = 0

plz help me??

Is "Z" and "z" two different variables? What are you solving for? I'll assume they're the same variable and that you're wanting to solve for z.

z^2 – 5*(1 + i)*z + 2 + 14*i = 0

There is no such z that will satisfy that equation.
• Oct 7th 2006, 09:32 AM
AfterShock
Disregard my previous post, I realized I was using the wrong data.

There are two solutions that will work for a z, such that that equation is equal to zero.

Try: 3 + i and 2 + 4*i

Notice that the simplified equation is:

z^2 - 5*z - 5*i*z + 2 + 14*i
• Oct 7th 2006, 09:38 AM
AfterShock
Since,

z^2 - 5*z - 5*i*z + 2 + 14*i = 0,

Factor it to get:

(z - 2 - 4*i)*(z - 3 - i) = 0

Then:

(z-2-4*i) = 0 AND
(z - 3 - i) = 0

For the first, z = 2 + 4*i
Second: z = 3 + i
• Oct 7th 2006, 09:50 AM
topsquark
Quote:

Originally Posted by AfterShock
Is "Z" and "z" two different variables? What are you solving for? I'll assume they're the same variable and that you're wanting to solve for z.

z^2 – 5*(1 + i)*z + 2 + 14*i = 0

There is no such z that will satisfy that equation.

AfterShock's factoring method is better than this one, but remember any quadratic with real or complex coefficients may be solved using the quadratic formula. The solution in this case is a bit of a mess (which is why I recommend factoring whenever possible) but will work.

-Dan