# 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,

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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)

Complex Numbers Loci in the Argand diagram E.g. 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 z is closer (or as close) to 2 than to -2j the direction from 2 to z is π/4: a half-line

De Moivres theorem Polar form: if z = x + yj has modulus r and argument θ, Multiplication and division and may need to add or subtract 2π to give principal argument –π < θ π

De Moivres theorem As a consequence of these two, Uses of de Moivres 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 IN F.BOOK

De Moivres theorem 2. cos n θ and sin n θ in terms of multiple angles and KNOW and KNOW

De Moivres theorem Example Express cos 3 θ in terms of cos 3θ and cos θ do this right! use: integration dont forget 2s

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

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

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

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

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

Questions: Winter 06

Examiners 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

Questions: Summer 06

Examiners 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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