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Stanford Math Circle January 21, 2007

Tatiana Shubin(shubin@math.sjsu.edu)

Complex Numbers

Some History

Let us try to solve the equation 4153 += xx .x = 4 is an obvious solution.Also, 0)14)(4(415 23 =++−=−− xxxxx yields 2 more solutions

32142 ±−=−±−=x .

If we draw the graphs of 3xy = and 415 += xy , we can see that they intersect at 3points.Notice also that any equation qpxx 233 += must have at least one solution!

Yet the Cardan-Tartaglia formula (Ars Magna, 1545)

3 323 32 pqqpqqx −−+−+=

gives33 12121212 −−+−+=x .

Around 1572, Bombelli guessed that 112123 −+=−+ ba ,

112123 −−=−− ba , and found a = 2, b = 1. Then

4)12()12( =−−+−+=x .

And thus the complex numbers really entered the scene of mathematics in 1572.But they remained somewhat “unreal” until Gauss (almost two and a half centuries later)introduced the idea of treating them as points on the plane.

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Definition A complex number is a point z =(x, y) in the Cartesian plane in which thex-axis is measured in ordinary (real) units and the y-axis is measured in a different(imaginary) unit i. The real number x is called the real part of z (x = Re(z)), and thereal number y is called the imaginary part (y = Im(z)). (Note: Both the real and theimaginary parts of z are real numbers.)

Operations on Complex Numbers

Algebra Geometry

Vector z = (x, y), w = (s, t)Addition: z + w = (x + s, y + t)

Scalar z = (x, y), k realMultiplication: kz = (kx, ky)

We can now think of complex numbers as vectors. More precisely, the set ofcomplex numbers can be thought of as equivalent to the real vector space R2.

Corollary: z = (x, y) = x(1, 0) + y(0, 1) = x + iy (Cartesian form of z).

Question: What is z – w ? (Algebraically and geometrically).

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Important Notions

Algebra Geometry

Modulus: 22 yxz +=

Conjugate: iyxz −=

Questions: 1. What is wz − ? (Algebraically and geometrically).

2. Is zz = ?

Complex Multiplication

We want 12 −=i , and we also want the usual algebraic rules to still hold: associative,commutative laws for both addition and multiplication, and also, the distributive law.So, if iyxz += , then ixyiyxiiz +−=+= )( .

Question: Geometrically, how can we get iz?

From the picture above, what can you conclude about the relationship of the vectors z andiz?

In general, how can we get wz, where itsw += ?

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Let P, Q, M, N, R, T denote the points of the complex plain that correspond to the pointsz, w, sz, tz, itz, )()( iztszzitswz +=+= , respectively.ROMT is a parallelogram, and since RON∠ is right, ROMT is a rectangle. Consider thetriangles TMO and QXO where X corresponds to the number s = s + i0. QXOTMO ∠≅∠

and OX

QX

s

t

zs

zt

sz

itz

OM

OR

OM

TM===== , and hence QXOTMO ΔΔ ~ .

Thus (1) QOXTOM ∠≅∠ , and

(2) Since zOXzsszMT === , so zwzOQOTwz === .

So to get wz we rotate z counterclockwise through the angle QOX∠ , then stretch it by

a factor of w .

Another Form of Complex Numbers

Complex numbers are points of a plane, so we need 2 coordinates to denote ( = to locate)them. But we can use different ways to choose these coordinates. In particular, we canuse polar coordinates: ),( θrz = , where zr = , and )2(mod)arg( πθ z= .

)sin(cos θθ irz +=

Notice: We have seen that if ),(),,( µθ wwzz == , then

).,( µθ += wzzw

Hence wzzw = (multiply moduli);

arg(zw) = arg(z) + arg(w) (add arguments).

Cartesian Form Polar Form

Complexmultiplication: θθθ iezizzitswiyxz =+=+=+= )sin(cos,

µµµ iewiww =+= )sin(cos

)()( xtysiytxszw ++−= ))sin()(cos( µθµθ +++= iwzzw

)( µθ += iewz

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Problems1. Use polar form to show that every complex number 0≠z has two square roots. Hint:What happens (geometrically!) when a complex number is squared? Therefore, what isnecessary to “unsquare” (that is to take the square root) of a number (geometrically)?

2. In Cartesian form, one square root of iyx + when 0≥y is

22

2222 xyxi

xyxiyx

−++

++=+ .

Verify this. What is the other root? What if 0<y ?

3. Find all square roots of these complex numbers:(a) 1− (c) i (e) i− (g) i22 +

(b) 4− (d) 2I (f) 31 i−− (h) 4

Properties of conjugates and moduli

1. wzwz +≤+ (Triangle Inequality)

2. wzzw =

3. ( ) 2www =

4. ( ) ( )( )wzzw =

5. wzwzandwzwz −=−+=+ ,

6. w

z

w

z=

7. ( )( )wz

w

z=

Notice:( )( )

( )∈==

2

1

w

w

ww

w

wC (if 0≠w )

Fundamental Theorem of Algebra

(Proved in 1799 by Gauss, 21 years old at the time).

A polynomial equation of degree n>1 with complex coefficients has a solution in the setof all complex numbers, C. (And thus C is algebraically closed).

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Corollary: A polynomial equation over C of degree n has exactly n roots in C,counting multiplicities.

Some Other Important Properties

1. )sin(cos θθ ninzznn += (where )sin(cos θθ izz += )

(De Moivre’s Theorem).

2. This theorem can be used to find all nth roots of complex numbers.

Examples: Find all solutions of the given equations.1. 83 =z2. 15 =z

ADDITIONAL EXERCISES AND PROBLEMS

1. Perform the indicated operations, and reduce each of the following numbers to theform ,iyx + where ∈yx, R.

(a) )3)(2)(1( iii −−− (c))43(

)34(

i

i

−

+

(b) 6)3( i+ (d)z

z

+

−

5

5, where iz 34 +=

2. Find necessary and sufficient conditions for the three numbers 321 ,, zandzz to

be the vertices of an equilateral triangle.

3. Let ∈+= baibaz ,, R. Find conditions on a and b such that

(a) 4z is real(b) 4z is purely imaginary.

4. Find the absolute value (modulus) of(a) i23+ (c) )34)(32)(1( iiii +−+−

(b) 31 i+− (d))32)(21(

)3(

ii

i

−+−

−

5. If a, b, and n are positive integers, prove that there exist integers x and y suchthat 2222 )( yxba n +=+ .

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6. (a) Show that if z is a root of a polynomial equation with real coefficients(i.e., all the coefficients are real numbers), then z is also a root.

(b) Show that a polynomial equation with real coefficients and odd degree musthave at least one real root.

7. Solve the equations.(a) 03 =− iz (c) 0325 =+z(b) 014 =+z (d) 016 =−z

8. Find the smallest positive integers m and n satisfying nm ii )1()31( −=+ .

9. Suppose that nAAA ...21 is a regular polygon inscribed in a circle of radius r and

center O. Let P be a point on 1OA extended beyond 1A . Show that

∏=

−=n

k

nnk rOPPA

1

.

10. Given a point P on the circumference of a unit circle and the vertices nAAA ,...,, 21

of an inscribed regular polygon of n sides, prove that PA12+PA2

2+…+PAn2 is a

constant.

11. Let A0, A1, A2, A3, A4 divide a unit circle into five equal parts. Prove that thechords A0A1 and A0A2 satisfy the equation ( (A0A1)(A0A2))

2 = 5.

12. (a) If z1 + z2 + z3 = 0, and 1z = 2z = 3z = 1, show that these three complex

numbers are the vertices of an equilateral triangle inscribed in the unit circle.(c) Can you extend the previous result to the case of four complex numbers? n complex numbers?