Chap
.
13
Complex Numbers and Functions
Sec
.
13
.
1
Complex Numbers
.
Complex Plane
Make sure that you really understand what is done in Eq. (7). Make also sure that you understand (8), the
formulas for going from complex back to real.
Problem Set
.
13
.
1
.
Page 606
1
.
Powers of
i
.
i
2
u003d
u2212
1 and
1
i
u003d
u2212
i
will be used quite frequently. A formal derivation of
i
2
u003d
u2212
1 from the
multiplication formula is shown in the text. 1/
i
u003d
u2212
i
/
i
ue0a2
u2212
i
ue0a3
u003d
u2212
i
/1
u003d
u2212
i
follows from (7).
7
.
Complex arithmetic
.
Here you first multiply complex numbers by real numbers, then add complex
numbers, and finally square the result; note that the result of such operations on complex numbers can
very well be a real number.
5
z
1
is obtained by multiplying the real part 2 by 5 and the imaginary part 3 by 5,
5
z
1
u003d
5
ue0a2
2
u002b
3
i
ue0a3
u003d
5
u2219
2
u002b
5
u2219
3
i
u003d
10
u002b
15
i
.
Similarly,
3
z
2
u003d
3
ue0a2
4
u2212
5
i
ue0a3
u003d
12
u2212
15
i
.
Now you sum the results (the first two pairs of parentheses will not be needed; why?)
5
z
1
u002b
3
z
2
u003d
ue0a2
10
u002b
15
i
ue0a3
u002b
ue0a2
12
u2212
15
i
ue0a3
u003d
10
u002b
12
u002b
ue0a2
15
u2212
15
ue0a3
i
u003d
22
and square, 22
2
u003d
484.
9
.
Division
,
real part
.
Do the division first and then take the real part. In the division you have a choice.
You can divide first and then take the square of the result, or you square first and then divide (take the
reciprocal). Try both ways, and you will see that the second way is simpler because 1/
z
1
gives fractions,
which you then would have to square. So do
z
1
2
u003d
ue0a2
2
u002b
3
i
ue0a3
2
u003d
4
u002b
2
u2219
2
u2219
3
i
u2212
9
u003d
u2212
5
u002b
12
i
.
Then do the division, using the recipe (7),
1
z
1
2
u003d
u2212
5
u2212
12
i
ue0a2
u2212
5
u002b
12
i
ue0a3ue0a2
u2212
5
u2212
12
i
ue0a3
u003d
u2212
5
u2212
12
i
25
u002b
144
u003d
u2212
5
u2212
12
i
169
.
Now take the real part
u2212
5/169.
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Chap. 13
Complex Numbers and Functions
129
19
.
Division
,
conjugate
.
You can do the division for the given expression. Or you can calculate 1/
z
u0304
and
square the result. In both cases, taking the real part is the last step.
In the first way, calculate
1
z
u0304
2
u003d
z
2
z
2
z
u0304
2
u003d
x
2
u002b
2
ixy
u2212
y
2
ue0a2
zz
u0304
ue0a3
2
u003d
x
2
u2212
y
2
u002b
2
ixy
ue0a2
x
2
u002b
y
2
ue0a3
2
and then take the real part to get
x
2
u2212
y
2
in the numerator and the denominator as before.
In the second way, you calculate
1
z
u0304
u003d
z
zz
u0304
u003d
x
u002b
iy
x
2
u002b
y
2
and then square the expression on the right, obtaining
x
2
u002b
2
ixy
u2212
y
2
ue0a2
x
2
u002b
y
2
ue0a3
2
,
of which you now take the real part as before.
Sec
.
13
.
2
Polar Form of Complex Numbers
.
Powers and Roots
The role of polar coordinates is more important in complex analysis than in calculus. Notably, they help
achieve a deeper understanding of multiplication and division. They also help in handling absolute values.
Now the polar angle of a complex number is determined only up to integer multiples of 2
u03c0
. Often this is
not essential, but where it matters, it can be handled by using the concept of the principal value (5).
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 Spring '12
 M.lee
 Complex Numbers, Cos, Complex number

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