ALGEBRA - solved problems
Imaginary and complex numbers
Imaginary numbers
 114 Reduce,  i15,  i26  and  i149.
Solutions:       i15 = i4 · 3 · i3 = i3 = - i,     i26 = i4 · 6 · i2 = i2 -1   and   i149 = i4 · 37 · i.
Complex numbers
Addition and subtraction of complex numbers
 115
Given are complex numbers,  z1 = -3 + 2i  and  z2 = 4 + 3i,  find  z1 + z2  and  z1 - z2.
Solutions:       z1 + z2 = (-3 + 2i) + (4 + 3i) = (-3 + 4) + (2 + 3)i = 1 + 5i
and             z1 - z2 = (4 + 3i) - (1 + 5i) = (4 - 1) + (3 - 5)i = 3 - 2i
Given addition and subtraction are shown in the complex plane in the figures below.
 z1 + z2 = (-3 + 2i) + (4 + 3i) = 1 + 5i z1 - z2 = (4 + 3i) - (1 + 5i) = 3 - 2i
Multiplication and division of complex numbers
 116
Given are complex numbers,  z1 = -3 + 2i  and  z2 = 4 + 3i, find  z1 ·  z2  and  z1 / z2.
Solutions:       z1 ·  z2 = (-3 + 2i) · (4 + 3i) = -3 · 4 + 2 · 4i + (-3) · 3i +  2 · 3 i2 = -18 - i
 and
 117
For what real number a the real part of the complex number equals 1.
 Solution:
 118
Evaluate the expression where  z = 1 - i.
 Solution:
Polar or trigonometric notation of complex numbers
 119 Given the complex number  z = 1 - Ö3 i,  express  z = x + yi  in the trigonometric form.
 Solution:   The modulus the argument
 the trigonometric form is
Multiplication and division of complex numbers in the polar form
If given    z1 = r1(cosj1 + isinj1)    and    z2 = r2(cosj2 + isinj2)
 then, z1 ·  z2 = r1 r2 · [cos(j1 + j2) + isin(j1 + j2)]
 and
Exponentiation and root extraction of complex numbers in the polar form - de Moivre's formula
We use the polar form for exponentiation and root extraction of complex numbers that are known as
 de Moivre's formulas, zn = rn · [cos (nj) + isin (nj)]
 and
 120
Calculate
 Solution:
 or in the polar form,
 and
since exponentiation with an integer exponent   zn = rn e inj = rn (cos nj + sin nj),
 then
 121
Compute
Solution:   Since the square root of a complex number is a complex number, then
and, two complex numbers are equal if their real parts are equal and their imaginary parts are equal, that is
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