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3.1 Complex number calculations

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Theory

Exercises

Contents:

Real and imaginary part
Addition and subtraction of complex numbers
Complex conjugate
Multiplication and division of complex numbers

Learning outcomes:

After this section, you will have learned to:

Simplify expressions that are constructed using complex numbers and the four arithmetic operations.
Solve first order complex number equations and simplify the answer.

Introduction

The real numbers represent a complete set of numbers in the sense that they "fill" the real-number axis. Despite this, the set of real numbers does not contain the solutions of all possible algebraic equations. In other words, there are equations of the type

\displaystyle a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0

that do not have a solution among the real numbers. For example, the equation \displaystyle x^2+1=0 has no real solution, since no real number satisfies \displaystyle x^2=-1. However, if we can imagine \displaystyle \sqrt{-1} as the number that satisfies the equation \displaystyle x^2=-1 and manipulate \displaystyle \sqrt{-1} just like any other number, it turns out that every algebraic equation does have solutions. The number \displaystyle \sqrt{-1}, however, is a fairly strange object. We cannot go out into the world and measure \displaystyle \sqrt{-1} anywhere, or find something that is numerically \displaystyle \sqrt{-1}. Nonetheless, this number turns out to be very useful in many applications of mathematics.

Example 1

If we would like to find out the sum of the roots (solutions) of the equation \displaystyle x^2-2x+2=0 we can first obtain the roots \displaystyle x_1=1+\sqrt{-1} and \displaystyle x_2=1-\sqrt{-1}. These roots contain \displaystyle \sqrt{-1}. If we allow ourselves to do calculations containing \displaystyle \sqrt{-1}, we see that the sum of \displaystyle x_1 and \displaystyle x_2 turns out to be \displaystyle 1+\sqrt{-1} + 1-\sqrt{-1} =2, which is an ordinary, familiar "real" number.

Even though the answer to the problem was a "real" number, we found the "imaginary" number \displaystyle \sqrt{-1} useful in solving it.

Definition of complex numbers

The terms "real" (for the ordinary, familiar positive and negative numbers, together with zero) and "imaginary" (for numbers like \displaystyle \sqrt{-1}) are actually pretty flawed in some ways (all numbers are abstract human inventions, it could be argued). Nonetheless, this terminology, which reflects the unease and scepticism with which the mathematical community once viewed numbers like \displaystyle \sqrt{-1}, has stuck.

The number \displaystyle \sqrt{-1} cannot be given an approximate decimal value, the way \displaystyle \sqrt{2} can; we have no choice but to write it as an unevaluated root (a surd). For simplicity and economy, we usually represent this number by the symbol \displaystyle i (sometimes \displaystyle j). It is sometimes called the imaginary unit, and any number of the form \displaystyle b\,i, where \displaystyle b is real, is known as an imaginary number. We go on to define a complex number as an object that can be written in the form

\displaystyle z=a+bi\,\mbox{,}

where \displaystyle a and \displaystyle b are real numbers, and \displaystyle i satisfies \displaystyle i^2=-1.

If \displaystyle a = 0 then the number is "purely imaginary". If \displaystyle b = 0 the number is real. We can see that the real numbers are a subset of the complex numbers. The set of complex numbers is designated by C.

For an arbitrary complex number one often uses the symbol \displaystyle z. If \displaystyle z=a+bi, where \displaystyle a and \displaystyle b are real, then \displaystyle a is the real part and \displaystyle b the imaginary part of \displaystyle z. One uses the following notation:

\displaystyle \begin{align*}a &= \mathop{\rm Re} z\,\mbox{,}\\ b&=\mathop{\rm Im} z\,\mbox{.}\end{align*}

When one calculates with complex numbers one treats them just like real numbers, but keeps track of the fact that \displaystyle i^2=-1.

Addition and subtraction

To add or subtract complex numbers one adds or subtracts the real and imaginary parts separately. If \displaystyle z=a+bi and \displaystyle w=c+di are two complex numbers then,

\displaystyle \begin{align*} z+w &= a+bi+c+di = a+c+(b+d)i\,\mbox{,}\\ z-w &= a+bi-(c+di) = a-c+(b-d)i\,\mbox{.}\end{align*}

Example 2

\displaystyle (3-5i)+(-4+i)=-1-4i
\displaystyle \bigl(\tfrac{1}{2}+2i\bigr)-\bigl(\tfrac{1}{6}+3i\bigr) = \tfrac{1}{3}-i
\displaystyle \frac{3+2i}{5}-\frac{3-i}{2} = \frac{6+4i}{10}-\frac{15-5i}{10} = \frac{-9+9i}{10} = -0\textrm{.}9 + 0\textrm{.}9i

Multiplication

Complex numbers are multiplied in the same way as ordinary real numbers or algebraic expressions, with the extra condition that \displaystyle i^2=-1. Generally one has for two complex numbers \displaystyle z=a+bi and \displaystyle w=c+di that

\displaystyle z\, w=(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i\,\mbox{.}

Example 3

\displaystyle 3(4-i)=12-3i
\displaystyle 2i(3-5i)=6i-10i^2=10+6i
\displaystyle (1+i)(2+3i)=2+3i+2i+3i^2=-1+5i
\displaystyle (3+2i)(3-2i)=3^2-(2i)^2=9-4i^2=13
\displaystyle (3+i)^2=3^2+2\times3i+i^2=8+6i
\displaystyle i^{12}=(i^2)^6=(-1)^6=1
\displaystyle i^{23}=i^{22}\times i=(i^2)^{11}\times i=(-1)^{11}i=-i

Complex conjugate

If \displaystyle z=a+bi then \displaystyle \overline{z} = a-bi is called the complex conjugate of \displaystyle z (the opposite is also true, that \displaystyle z is conjugate to \displaystyle \overline{z}). One obtains the relationships

\displaystyle \begin{align*} z+\overline{z} &= a + bi + a - bi = 2a = 2\, \mathop{\rm Re} z\,\mbox{,}\\ z-\overline{z} &= a + bi - (a - bi) = 2bi = 2i\, \mathop{\rm Im} z\,\mbox{,}\end{align*}

but most importantly, using the difference of two squares rule, one obtains

\displaystyle z\, \bar z = (a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2=a^2+b^2\,\mbox{,}

i.e. that the product of a complex number and its conjugate is always real.

Example 4

\displaystyle z=5+i\qquad then \displaystyle \quad\overline{z}=5-i\,.
\displaystyle z=-3-2i\qquad then \displaystyle \quad\overline{z} =-3+2i\,.
\displaystyle z=17\qquad then \displaystyle \quad\overline{z} =17\,.
\displaystyle z=i\qquad then \displaystyle \quad\overline{z} =-i\,.
\displaystyle z=-5i\qquad then \displaystyle \quad\overline{z} =5i\,.

Example 5

If \displaystyle z=4+3i one has
- \displaystyle z+\overline{z} = 4 + 3i + 4 -3i = 8
- \displaystyle z-\overline{z} = 6i
- \displaystyle z \, \overline{z} = 4^2-(3i)^2=16+9=25
If for \displaystyle z one has \displaystyle \mathop{\rm Re} z=-2 and \displaystyle \mathop{\rm Im} z=1, one gets
- \displaystyle z+\overline{z} = 2\,\mathop{\rm Re} z = -4
- \displaystyle z-\overline{z} = 2i\,\mathop{\rm Im} z = 2i
- \displaystyle z\, \overline{z} = (-2)^2+1^2=5

Division

For the division of two complex numbers one multiplies the numerator and denominator by the complex conjugate of the denominator, thus getting a denominator which is a real number. Then, both the real and imaginary parts of the (new) numerator are divided by this number (the new denominator). In general, if \displaystyle z=a+bi and \displaystyle w=c+di:

\displaystyle \frac{z}{w} = \frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}\,i

Example 6

\displaystyle \quad\frac{4+2i}{1+i} = \frac{(4+2i)(1-i)}{(1+i)(1-i)} = \frac{4-4i+2i-2i^2}{1-i^2} = \frac{6-2i}{2}=3-i
\displaystyle \quad\frac{25}{3-4i} = \frac{25(3+4i)}{(3-4i)(3+4i)} = \frac{25(3+4i)}{3^2-16i^2} = \frac{25(3+4i)}{25} = 3+4i
\displaystyle \quad\frac{3-2i}{i} = \frac{(3-2i)(-i)}{i(-i)} = \frac{-3i+2i^2}{-i^2} = \frac{-2-3i}{1} = -2-3i

Example 7

\displaystyle \quad\frac{2}{2-i}-\frac{i}{1+i} = \frac{2(2+i)}{(2-i)(2+i)} - \frac{i(1-i)}{(1+i)(1-i)} = \frac{4+2i}{5}-\frac{1+i}{2}
\displaystyle \quad\phantom{\frac{2}{2-i}-\frac{i}{1+i}}{} = \frac{8+4i}{10}-\frac{5+5i}{10} = \frac{3-i}{10}\vphantom{\Biggl(}
\displaystyle \quad\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}} = \frac{\dfrac{1-i}{1-i}-\dfrac{2}{1-i}}{\dfrac{2i(2+i)}{(2+i)} + \dfrac{i}{2+i}} = \frac{\dfrac{1-i-2}{1-i}}{\dfrac{4i+2i^2 + i}{2+i}} = \frac{\dfrac{-1-i}{1-i}}{\dfrac{-2+5i}{2+i}}
\displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-1-i}{1-i}\times \frac{2+i}{-2+5i} = \frac{(-1-i)(2+i)}{(1-i)(-2+5i)} = \frac{-2-i-2i-i^2}{-2+5i+2i-5i^2}\vphantom{\Biggl(}
\displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-1-3i}{3+7i} = \frac{(-1-3i)(3-7i)}{(3+7i)(3-7i)} = \frac{-3+7i-9i+21i^2}{3^2-49i^2}\vphantom{\Biggl(} \displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-24-2i}{58} = \frac{-12-i}{29}\vphantom{\Biggl(}

Example 8

Determine the real number \displaystyle a such that the expression \displaystyle \ \frac{2-3i}{2+ai}\ becomes real.

Multiply the numerator and denominator by the complex conjugate of the denominator so that the expression can be written with separate real and imaginary parts.

\displaystyle \frac{(2-3i)(2-ai)}{(2+ai)(2-ai)} = \frac{4-2ai-6i+3ai^2}{4-a^2i^2} = \frac{4-3a-(2a+6)i}{4+a^2}

If the expression is to be real , the imaginary part must be 0, ie.

\displaystyle 2a+6=0\quad\Leftrightarrow\quad a = -3\,\mbox{.}

Equations

If the two complex numbers \displaystyle z=a+bi and \displaystyle w=c+di are equal then it is not hard to show that both the real and imaginary parts must be equal (in other words, \displaystyle a=c and \displaystyle b=d). When you are looking for an unknown complex number \displaystyle z in an equation, you can either try to solve for the number \displaystyle z in the usual way, or insert \displaystyle z=a+bi in the equation and then compare the real and imaginary parts of the two sides of the equation with each other.

Example 9

Solve the equation \displaystyle 3z+1-i=z-3+7i.

Collect all \displaystyle z on the left-hand side by subtracting \displaystyle z from both sides

\displaystyle 2z+1-i = -3+7i
and now subtract \displaystyle 1-i

\displaystyle 2z = -4+8i\,\mbox{.}
This gives that \displaystyle \ z=\frac{-4+8i}{2}=-2+4i\,\mbox{.}
Solve the equation \displaystyle z(-1-i)=6-2i.

Divide both sides by \displaystyle -1-i in order to obtain \displaystyle z

\displaystyle z = \frac{6-2i}{-1-i} = \frac{(6-2i)(-1+i)}{(-1-i)(-1+i)} = \frac{-6+6i+2i-2i^2}{(-1)^2 -i^2} = \frac{-4+8i}{2} = -2+4i\,\mbox{.}
Solve the equation \displaystyle 3iz-2i=1-z.

Adding \displaystyle z and \displaystyle 2i to both sides gives

\displaystyle 3iz+z=1+2i\quad \Leftrightarrow \quad z(3i+1)=1+2i\,\mbox{.}

This gives

\displaystyle z = \frac{1+2i}{1+3i} = \frac{(1+2i)(1-3i)}{(1+3i)(1-3i)} = \frac{1-3i+2i-6i^2}{1-9i^2} = \frac{7-i}{10}\,\mbox{.}
Solve the equation \displaystyle 2z+1-i=\bar z +3 + 2i.

The equation contains both \displaystyle z as well as \displaystyle \overline{z} and therefore we assume \displaystyle z to be \displaystyle z=a+ib and solve the equation for \displaystyle a and \displaystyle b by equating the real and imaginary parts of both sides

\displaystyle 2(a+bi)+1-i=(a-bi)+3+2i

i.e.

\displaystyle (2a+1)+(2b-1)i=(a+3)+(2-b)i\,\mbox{,}

which gives

\displaystyle \left\{\begin{align*} 2a+1 &= a+3\\ 2b-1 &= 2-b\end{align*}\right.\quad\Leftrightarrow\quad\left\{\begin{align*} a &= 2\\ b &= 1\end{align*}\right.\,\mbox{.}
The answer is therefore, \displaystyle z=2+i.

Study advice

Keep in mind that:

Calculations with complex numbers are done in the same way as with ordinary numbers with the additional information that \displaystyle i^2=-1.

Quotients of complex numbers are simplified by multiplying the numerator and denominator by the complex conjugate of the denominator.

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1. Derivatives

2. Integrals

3. Complex numbers

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