FP1 Complex Numbers

Given that

z = ( 2 - i ) & z² = ( 3 - 4i )

Find the two roots of the following equation.

( z + i )² = ( 3 - 4i )

Thanks in advance.

Try rewriting it as

(z+i)^{2} = (2-i)^{2}

Why should that be so?

( z + i )² ≠ ( 2 - i )²

Can you explain why it should be writen like that?

Thanks

Bengaltiger

Why should that be so?

( z + i )² ≠ ( 2 - i )²

Can you explain why it should be writen like that?

Thanks

From the equation,

(z + i)² = (3 - 4i), which equals z². Sub in z, and you get (2 - i)²
∴(z + i)² = (2 - i)²

Bengaltiger

Given that

z = ( 2 - i ) & z² = ( 3 - 4i )

Find the two roots of the following equation.

( z + i )² = ( 3 - 4i )

z = ( 2 - i )
z² = ( 2 - i )²

( 3 - 4i ) = ( 2 - i )²

( z + i )² = ( 2 - i )²
z² + 2zi - 1 = 4 - 4i - 1
z² + 2zi + 4(i - 1) = 0

solve for z.

ok got that far, but how to solve it now?

quadratic formula

Using the formula is quite tricky on this. Quite hard to simplfy.

Why not (z+i)² = (2-i)²

=> z+i = ± (2-i)

=> z = -i ± (2-i)

=> z = 2-2i or z = -2

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