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Complex Numbers

Complex Numbers . Ideas from Further Pure 1 What complex numbers are The idea of real and imaginary parts if z = 4 + 5 j , Re( z ) = 4, Im( z ) = 5 How to add, subtract, multiply and (especially) divide complex numbers The Argand diagram Modulus (easy) and argument (often wrong).

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Complex Numbers

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  1. Complex Numbers Ideas from Further Pure 1 • What complex numbers are • The idea of real and imaginary parts • if z = 4 + 5j, Re(z) = 4, Im(z) = 5 • How to add, subtract, multiply and (especially) divide complex numbers • The Argand diagram • Modulus (easy) and argument (often wrong)

  2. Complex Numbers • Loci in the Argand diagram • E.g. z is closer (or as close) to 2 than to -2j the direction from 2 to z is π/4: a half-line • Equations: a polynomial equation of degree n has precisely n (real or complex) roots (some may be repeats) • The complex roots of polynomial equations with real coefficients occur in conjugate pairs

  3. De Moivre’s theorem • Polar form: if z = x + yj has modulus r and argument θ, • Multiplication and division and and may need to add or subtract 2π to give “principal argument” –π < θ ≤ π

  4. De Moivre’s theorem • As a consequence of these two, IN F.BOOK Uses of de Moivre’s theorem 1. cos nθ and sin nθ in terms of powers by de M useful abbreviation by Binomial • Then equate real and imaginary parts • May need to use

  5. De Moivre’s theorem 2. cosnθ and sinnθ in terms of multiple angles   and KNOW   and KNOW

  6. De Moivre’s theorem • Example • Express cos3θ in terms of cos 3θ and cos θ do this right! don’t forget 2s  use: integration

  7. Complex exponents Use: Summing series Example: Call this C, define S = Then C + jS = This is a G.P. with a = 1, r = which we have to simplify to find C Sum to infinity =

  8. Complex exponents is Complex conjugate of C + jS = Now so C + jS = C is the real part: C =

  9. Complex roots • We want to find the nth roots of a complex number w • Suppose Then • On the Argand diagram: • roots lie on a circle, radius • they are separated by 2π/n so form an n-sided polygon inscribed in the circle

  10. Complex roots • Example: Find the cube roots of 8 + 8j Cube roots have modulus Arguments so principal arguments are so roots are

  11. Complex loci • There is one more technique in FP2 The vector from 2 to z is π/4 ahead of the vector from -5 to z so locus is arc of circle, endpoints -5 and 2

  12. Questions: Winter 06

  13. Examiner’s Report (i) Found difficult! Modulus much better than argument (ii) This was done well, but sometimes proofs lacked sufficient detail (iii) Not done well: some candidates appeared not to have been taught this Link with part (ii) not recognised Some had sums to n terms but went on to let n tend to infinity

  14. Questions: Summer 06

  15. Examiner’s Report (a) (i) Most could write this down (ii) This exact question is in the book! A lot got tangled up, or left out the powers of 2 (b) (i) Little trouble (ii) Efficiently done, although some gave arguments outside the range (iii) Many did not see connection with (ii) and started again

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