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ALGEBRA
- solved problems |
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Imaginary
and complex numbers
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Imaginary
numbers
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114. |
Reduce, i15,
i26
and i149.
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Solutions:
i15
= i4
· 3 · i3 = i3
= -
i,
i26
= i4
· 6 · i2 = i2
= -1
and i149
= i4
· 37 · i = i.
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Complex numbers
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Addition and
subtraction of complex numbers
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Given
are complex numbers, z1
= -3 +
2i and
z2
= 4 + 3i,
find z1 +
z2 and
z1
-
z2.
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Solutions:
z1 +
z2
= (-3 +
2i) + (4 + 3i) =
(-3 +
4) + (2 + 3)i = 1 + 5i
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and
z1
-
z2
= (4 + 3i) -
(1 + 5i) =
(4 -
1) + (3 -
5)i =
3 -
2i
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Given
addition and subtraction are shown in the complex plane in
the figures below.
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z1 +
z2
=
(-3 +
2i) + (4 + 3i) = 1 + 5i |
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z1
-
z2
=
(4 + 3i) -
(1 + 5i) = 3 -
2i |
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Multiplication and
division of
complex numbers
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Given
are complex numbers, z1
= -3 +
2i and
z2
= 4 + 3i,
find z1 ·
z2
and z1
/
z2.
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Solutions:
z1 ·
z2
= (-3 +
2i) · (4 + 3i) = -3 ·
4 + 2 · 4i +
(-3)
· 3i + 2 · 3 i2
= -18
-
i
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and |
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For what real
number
a the real part of the complex number |
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equals
1. |
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Solution: |
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Evaluate the
expression
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where
z = 1 -
i. |
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Solution: |
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Polar or trigonometric
notation of complex numbers
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119. |
Given
the complex number z
= 1
-
Ö3
i, express
z
= x
+
yi
in the trigonometric form.
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Solution:
The modulus |
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the argument |
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the trigonometric form is
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Multiplication
and division of complex numbers in the polar form
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If
given z1
= r1(cosj1
+ isinj1)
and z2
= r2(cosj2
+ isinj2)
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then, |
z1
· z2 = r1
r2 · [cos(j1
+ j2)
+ isin(j1
+ j2)] |
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and |
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Exponentiation
and root extraction of complex numbers in the polar form - de
Moivre's formula
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We use the polar form
for exponentiation and root extraction of complex numbers that
are known as
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de Moivre's formulas, |
zn
= rn · [cos
(nj)
+ isin (nj)] |
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and |
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Calculate
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Solution: |
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or
in the polar form, |
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and |
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since
exponentiation with an integer exponent zn
= rn e inj
= rn (cos nj
+ sin nj),
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then |
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Compute
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Solution:
Since the square root of a
complex number is a complex number, then
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and, two
complex numbers are equal if their real parts are equal and
their imaginary parts are equal, that is
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Solved
problems contents |
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