A brief note on The Argand Plane and Quadratic Equations . Read it before a test. Try each question at the end before you open its answer.
Conjugate and modulus For z = a + i b z = a + ib z = a + ib , the conjugate is z ˉ = a − i b \bar z = a - ib z ˉ = a − ib and the modulus is ∣ z ∣ = a 2 + b 2 |z| = \sqrt{a^2 + b^2} ∣ z ∣ = a 2 + b 2 . z z ˉ = ∣ z ∣ 2 z\bar z = |z|^2 z z ˉ = ∣ z ∣ 2 , z + z ˉ = 2 a z + \bar z = 2a z + z ˉ = 2 a and z − z ˉ = 2 i b z - \bar z = 2ib z − z ˉ = 2 ib . z 1 + z 2 ‾ = z ˉ 1 + z ˉ 2 \overline{z_1 + z_2} = \bar z_1 + \bar z_2 z 1 + z 2 = z ˉ 1 + z ˉ 2 and z 1 z 2 ‾ = z ˉ 1 z ˉ 2 \overline{z_1 z_2} = \bar z_1 \bar z_2 z 1 z 2 = z ˉ 1 z ˉ 2 . ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ |z_1 z_2| = |z_1||z_2| ∣ z 1 z 2 ∣ = ∣ z 1 ∣∣ z 2 ∣ and ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ \left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|} z 2 z 1 = ∣ z 2 ∣ ∣ z 1 ∣ . ∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣ |z_1 + z_2| \le |z_1| + |z_2| ∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣ , usually with a strict inequality. Dividing Multiply the top and the bottom by the conjugate of the bottom: z 1 z 2 = z 1 z ˉ 2 ∣ z 2 ∣ 2 \dfrac{z_1}{z_2} = \dfrac{z_1 \bar z_2}{|z_2|^2} z 2 z 1 = ∣ z 2 ∣ 2 z 1 z ˉ 2 . The inverse of z ≠ 0 z \ne 0 z = 0 is z − 1 = z ˉ ∣ z ∣ 2 z^{-1} = \dfrac{\bar z}{|z|^2} z − 1 = ∣ z ∣ 2 z ˉ . The Argand plane Idea On the plane a + i b a + ib a + ib the point ( a , b ) (a, b) ( a , b ) real numbers the real axis, across i b ib ib the imaginary axis, up $ z $ the distance from the origin z ˉ \bar z z ˉ the mirror image in the real axis − z -z − z the point opposite z z z through the origin $ z_1 - z_2 $ the distance between z 1 z_1 z 1 and z 2 z_2 z 2
Quadratic equations For a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 with real coefficients and D = b 2 − 4 a c < 0 D = b^2 - 4ac < 0 D = b 2 − 4 a c < 0 , the roots are − b ± i − D 2 a \dfrac{-b \pm i\sqrt{-D}}{2a} 2 a − b ± i − D . The roots are a conjugate pair, mirror images in the real axis. Their sum is − b a -\dfrac{b}{a} − a b , and their product is c a = ∣ α ∣ 2 \dfrac{c}{a} = |\alpha|^2 a c = ∣ α ∣ 2 . Square roots To find a + i b \sqrt{a + ib} a + ib , write ( x + i y ) 2 = a + i b (x + iy)^2 = a + ib ( x + i y ) 2 = a + ib . Then:
x 2 − y 2 = a x^2 - y^2 = a x 2 − y 2 = a and 2 x y = b 2xy = b 2 x y = b ; x 2 + y 2 = a 2 + b 2 x^2 + y^2 = \sqrt{a^2 + b^2} x 2 + y 2 = a 2 + b 2 ; x x x and y y y have the same sign when b > 0 b > 0 b > 0 , and opposite signs when b < 0 b < 0 b < 0 . A worked example Solve x 2 − 6 x + 34 = 0 x^2 - 6x + 34 = 0 x 2 − 6 x + 34 = 0 .
Step 1. The discriminant. D = 36 − 136 = − 100 D = 36 - 136 = -100 D = 36 − 136 = − 100 .
Step 2. The roots. x = 6 ± 10 i 2 = 3 ± 5 i x = \dfrac{6 \pm 10i}{2} = 3 \pm 5i x = 2 6 ± 10 i = 3 ± 5 i .
Step 3. Check. The sum is 6 6 6 , and the product is 9 + 25 = 34 9 + 25 = 34 9 + 25 = 34 .
Practice with new numbers
A program to try The program solves the worked example with cmath and checks that the roots are conjugates with equal moduli.
Run it. Is the product of the roots 34 34 34 ?
1. Change c to 13. What are the roots now, and their modulus?
Mistakes to avoid The wrong sign in the conjugate. Only the imaginary part changes sign. ∣ a + i b ∣ |a + ib| ∣ a + ib ∣ as a + b a + b a + b . It is a 2 + b 2 \sqrt{a^2 + b^2} a 2 + b 2 . Distance as ∣ z 1 ∣ − ∣ z 2 ∣ |z_1| - |z_2| ∣ z 1 ∣ − ∣ z 2 ∣ . It is ∣ z 1 − z 2 ∣ |z_1 - z_2| ∣ z 1 − z 2 ∣ . No solution for a negative D D D . There are two complex roots. − 36 \sqrt{-36} − 36 as − 6 -6 − 6 . It is 6 i 6i 6 i . Check yourself Write 4 + 3 i 2 − i \dfrac{4 + 3i}{2 - i} 2 − i 4 + 3 i in the form a + i b a + ib a + ib . Stuck? Show a hint Show the answerWhat is the modulus of ( 1 + i ) 6 (1 + i)^6 ( 1 + i ) 6 ? Stuck? Show a hint Show the answerThe points z z z and z ˉ \bar z z ˉ are 8 8 8 apart. What is the imaginary part of z z z , if it is positive? Stuck? Show a hint Show the answerSolve x 2 + 6 x + 13 = 0 x^2 + 6x + 13 = 0 x 2 + 6 x + 13 = 0 . Stuck? Show a hint Show the answerFind the square roots of − 21 − 20 i -21 - 20i − 21 − 20 i . Stuck? Show a hint Show the answer