Giải phương trình sau
z^2 − (3 − 2i)z + 5 − 5i = 0;

There are several ways to solve a quadratic equation. I'll complete the square:

z ² - (3 - 2i ) z + 5 - 5i = 0

(z - (3 - 2i )/2)² - ((3 - 2i )/2)² + 5 - 5i = 0

(z - (3 - 2i )/2)² - (3 - 2i )²/4 + 5 - 5i = 0

(z - (3 - 2i )/2)² = (3 - 2i )²/4 - 5 + 5i

(z - (3 - 2i )/2)² = (3² - 2×3×(-2i ) + (-2i )²)/4 - 5 + 5i

(z - (3 - 2i )/2)² = (5 - 12i )/4 - 5 + 5i

(z - (3 - 2i )/2)² = (5 - 12i - 20 + 20i )/4

(z - (3 - 2i )/2)² = (-15 + 8i )/4

Let w = (-15 + 8i )/4. Then write w = |w| exp(i arg(w)), where

|w| = √((-15/4)² + 2²) = 17/4

arg(w) = π - arctan(8/15)

The square roots of w are then

√w = √(|w|) exp(i (arg(w) + 2nπ)/2)

where n in the set {0, 1}.

Taking the square root of both sides gives

z - (3 - 2i )/2 = √w

z = (3 - 2i )/2 + √w

and the two solutions can be simplified to

z = √17/4 + √17 i

z = -√17/4 - √17 i

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: