Complex Numbers

Complex numbers arise naturally when solving quadratic equations. We know that the solutions to x21=0 x^2 -1 =0 are x=1,1 x = 1, -1 . What are the solutions to x2+1=0 x^2 + 1 = 0 ?

i i is often used to denote the imaginary unit, which satisfies the equation i2=1 i^2 = -1 . i i and i -i will be the roots to the equation x2+1=0 x^2 + 1 = 0 . With this symbol, we can extend the real numbers to obtain the set of complex numbers, which are of the form z=a+bi z= a+ bi , where a a and b b are real numbers.

We say that for a complex number z=a+bi z= a+bi , it has real part a a , denoted as Re(z) Re(z) , and imaginary part b b , denoted as Im(z) Im (z) . Take note that the imaginary part is a real number. Let's see how the usual arithmetic operations work:

1) Addition:        

(a+bi)+(c+di)=(a+c)+(b+d)i (a+bi) + (c+di) = (a+c) + (b+d)i .

2) Subtraction:  

(a+bi)(c+di)=(ac)+(bd)i (a+bi) - (c+di) = (a-c) + (b-d) i .

3) Multiplication:  

(a+bi)×(c+di)=ac+adi+bci+bdi2=(acbd)+(ad+bc)i (a+bi)\times (c+di) = ac + adi + bci + bdi^2 = (ac-bd) + (ad+bc) i .

Division becomes slightly tricky, because we only know how to divide by a real number. Rewriting a+bic+di \frac {a+bi} { c+di} as ac+di+bic+di \frac {a}{c+di} + \frac {bi}{c+di} isn't very helpful. If only we could make the denominator a real number ... To do so, we introduce the idea of a conjugate:

The conjugate of the complex number z=a+bi z = a+bi is z=abi \overline{z} = a - bi . Notice that the conjugate of the conjugate is the identity, i.e. z=z \overline{ \overline{z}} = z .

We now have:

zz=(a+bi)(abi)=[a×ab×(b)]+[a×(b)+b×a]i=a2+b2 z \overline{z} = (a+bi)(a-bi) = [ a \times a - b \times (-b) ] + [a \times (-b) + b \times a]i = a^2 + b^2

using our nifty multiplication formula. This gives us a real non-negative value. Now, we introduce the idea of a Norm and absolute value:

The norm of the complex number z=a+bi z = a+bi is N(z)=zz=a2+b2 N( z ) = z \overline{z} = a^2 + b^2 .The absolute value of the complex number is the positive square root of the norm, and is given by a+bi=a2+b2 \left| a+bi \right| = \sqrt{a^2+b^2}.

With this, we have the following:

4) Division:      If c+di c+di is non-zero, then a+bic+di=(a+bi)(cdi)(c+di)(cdi)=(acbd)+(ad+bc)ic2+d2 \frac {a+bi} {c+di} = \frac {(a+bi)(c-di)}{(c+di)(c-di)} = \frac {(ac-bd) + (-ad+bc)i}{c^2 + d^2 } .

z=x+iy z = x + iy is known as the Rectangular Form of the complex number. In an upcoming post, we will study the Polar Form of complex numbers.

Worked Examples

1. If a a and b b are real numbers such that a+bi=0 a + bi = 0 , then a=0 a = 0 and b=0 b = 0 .

This should be obvious to you, but let's show it. If a+bi=0 a + bi = 0 , then a=bi a = - bi . Squaring both sides, we get a2=(bi)2=b2i2=b2 a^2 = (-bi)^2 = b^2 i^2 = - b^2 . By non-negativity of squares, we have 0a2=b20 0 \leq a^2 = -b^2 \leq 0 , which implies that equality must hold throughout. Thus, 0=a2=b2 0 = a^2 = -b^2 , which gives a=0,b=0 a = 0, b= 0 .

Corollary: This allows us to compare coefficients of real and imaginary parts. In particular, if z=w z = w , then we must have Re(z)=Re(w),Im(z)=Im(w) Re(z) = Re(w), Im(z) = Im(w) .

 

2. With real numbers, we are familiar with the concept of reciprocals. For example, 2 and 12 \frac {1}{2} are reciprocals of each other because 2×12=1 2 \times \frac {1}{2} = 1 . What is the reciprocal of a non-zero complex number z=a+bi z = a+ bi ?

Solution 1: Since (a,b)(0,0) (a, b) \neq (0, 0) , z0 \left \| z \right \| \neq 0 . We seek the value of 1z \frac {1}{z} so will use the division operation, to obtain that 1z=1a+bi=abi(a+bi)(abi)=abia2+b2 \frac {1}{z} = \frac {1}{a+bi} = \frac {a-bi}{(a+bi) (a-bi)} = \frac {a-bi} {a^2 + b^2 } .

Solution 2: Using the language of norm and conjugates, we can express this directly as:1z=zzz=zN(z) \frac {1}{z} = \frac {\overline{z} } { z \overline{z} } = \frac {\overline{z} } { N( z ) } .

 

3. Verify that conjugation distributes over multiplication and division. Specifically, show that z×w=z×w \overline{z \times w} = \overline{z} \times \overline{w} and (zw)=zw \overline{ \left( \frac {z}{w} \right)} = \frac {\overline{z}} {\overline{w} } .

Let z=a+bi,w=c+di z = a + bi, w = c + di then z×w=(acbd)+(ad+bc)i=(acbd)(ad+bc)i, \begin{aligned} \overline { z \times w} &= \overline { (ac-bd) + (ad+bc) i } \\ &= (ac-bd) - (ad+bc) i, \end{aligned} while z×w=(abi)×(cdi)=[(ac+(bi)(di)]+[a(d)+(b)c]i=(acbd)(ad+bc)i. \begin{aligned} \overline{z} \times \overline{w} &= (a-bi) \times (c-di) \\ &= [( ac+ (-bi)(-di) ] + [a(-d) + (-b) c] i \\ &= (ac-bd) - (ad+bc) i. \end{aligned} Hence they are the same.

As for the division, let's use the language of norm and conjugates that we've learned. You can also do this using a,b,c,d a, b, c, d as above.

(zw)=(z×ww×w) multiplying by conjugates=(z×wN(w))definition of Norm=z×wN(w)since Norm is a real value=z×wˉN(w)since conjugation distributes over multiplication=z×wN(w)property of conjugation=z×ww×w definition of Norm=zw cancellation \begin{aligned} \overline{ \left( \frac {z}{w} \right)} & = \overline{ \left( \frac {z \times \overline{w}} {w \times \overline{w}} \right) } & & \mbox{ multiplying by conjugates} \\ & = \overline{ \left(\frac {z \times \overline{w}} {N( w )} \right) }& &\mbox{definition of Norm} \\ & =\frac { \overline{ z \times \overline{w}} } { N(w )} & &\mbox{since Norm is a real value}\\ & = \frac {\overline{z} \times \overline{\bar{w}}}{N( w )}&& \mbox{since conjugation distributes over multiplication} \\ & = \frac {\overline{z} \times w }{N( w ) } && \mbox{property of conjugation} \\ & = \frac {\overline{z} \times w} { \overline{w} \times w}& &\mbox{ definition of Norm} \\ & = \frac {\overline{z}} {\overline{w} }& &\mbox{ cancellation} \\ \end{aligned}

You should easily verify that conjugation distributes over addition and subtraction.

Note by Arron Kau
5 years, 6 months ago

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