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

plz help me??

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- Oct 7th 2006, 06:07 AMiceman1in 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 AMAfterShock
- Oct 7th 2006, 09:32 AMAfterShock
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 AMAfterShock
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 AMtopsquark
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