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Question from Ash, a student:

z^2-(6+2i)z+(8+6i)=0

Solve for Z

thanks for the help

we have two responses for you

Hi Ash,

I didn't see how to factor this so I decided to try the general quadratic. This involves the root of b2 - 4ac which I calculated to be

b2 - 4ac = (6 + 2i)2 - 4 (8 + 61) = 36 + 24i -4 - 32 - 24i = 0.

so the only solution is -b/2a = (6 +2 i)/2 = 3 + i.

Now that you know the answer it's easy to check. Expand [z - (3 + i)]2 and check that you get
z2 - (6+2i)z + (8+6i)

Penny

 

Hi Ash.

Try completing the square by taking the c term to the other side and using half the b term:

You have z2 + bz + c = 0, so to complete the square ( z + b/2 )2 = -c + (b/2)2.

Thus [ z - (3+i) ]2 = - (8+6i) + (3 + i)2

Solve for z.

Stephen La Rocque.

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