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Transcript
FP3: Complex Numbers De Moivre’s theorem Lesson objectives.: * to understand de Moivre’s theorem for positive and negative powers * to find the square roots of a complex number using this theorem. Suppose z is a complex number with modulus r and argument θ, i.e. z r (cos i sin ) OR, using shorthand notation, z = [r, θ]. Then z2 z3 z4 This suggests that z n De Moivre’s theorem states that if z r (cos i sin ) then zn This result applies for any positive or negative power n. Note: We will be able to prove this result more formally next lesson. Example: z 3 i . a) Find z 2 and z3 . b) Find z8 and z 3 Solution: First write z in modulus-argument form: z=[ , (Note: we should write both of these in exact form if we can) So z 2 [ , ] and z3 [ , ] ] The Cartesian versions of this complex numbers are: z2 and z3 We can show these complex numbers on an Argand diagram: Im 8 6 4 2 -8 -4 0 -2 -2 -4 -8 b) Likewise: z8 [ ] , z 3 [ , ] Example 2: w 0.8 0.6i . Find w9 . Solution: In mod-arg form: w = [ Therefore, w9 [ So w9 , , ]. ]. 2 4 6 8 Real The Argand diagram in this case is as follows: Im 1 0.5 -1 0 -0.5 Real 1 0.5 -0.5 -1 Note: Accuracy is important in this work – try to use exact values for the modulus and argument where possible. Finding the square root of a complex number In FP1, we considered one way to find the square roots of a complex number. There is an alternative way to find the roots using de Moivre’s theorem. Example: Find the square roots of 5 – 12i, giving your answers in the form a + bi. Solution: Write 5 – 12i in modulus-argument form: 5 – 12i = [13, -1.1760052] Using de Moivre’s theorem: Therefore 5 12i (we want to keep accuracy) 5 12i (5 12i)1/ 2 131/ 2 , 12 1.1760052 13, 0.5880026 So: 5 12i 13(cos(0.5880026) i sin(0.5880026)) = 13(0.83205 0.55470 i) = 3 2i But… there is a second square root. To find this other root we find an equivalent form for 5 – 12i by adding 2π to the argument (this gives an equivalent angle). So we also have: 5 – 12i = [13, -1.1760052 + 2π] = [13, 5.107180] Using the above method: Therefore: 5 12i 131/ 2 , 12 5.107180 13, 2.553590 5 12i 13(cos 2.553590 i sin 2.553590) 3 2i Note: If one square root is a + bi, the other will be –a – bi. But we will need to extend the above method in order to deal with cube roots, fourth roots etc. Example 2: Find the square roots of 1 + i. Solution: 1 + i = [ So: (1 i)1/ 2 [ , , ] OR [ ]=[ , , ]=[ ] OR [ , ]. , ]=[ Therefore (1 i)1/ 2 OR (1 i)1/ 2 Past examination question: Use de Moivre’s theorem to find the value of (1 2i)10 . Solution: Let z = 1 – 2i. Corrected version: Converting z to mod-arg form gives z = [2.24, -1.11]. Therefore z10 2.2410 , (1.11)10 3180.4, 2.84 So (1 2i)20 = 3180.4(sin2.84 + icos2.84) = 3180.4(0.297 + i(-0.955)) = 944.6 – 3037.3i , ].