Chapter 4: Complex Numbers and Quadratic Equations

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Writing the given equation in standard form:

$$4x + i(3x - y) = 3 + i(-6)$$

Equating the real parts: $$4x = 3$$, so $$x = \dfrac{3}{4}$$.

Equating the imaginary parts: $$3x - y = -6$$.

Substituting $$x = \dfrac{3}{4}$$:

$$3 \cdot \dfrac{3}{4} - y = -6$$

$$\dfrac{9}{4} - y = -6 \implies y = \dfrac{9}{4} + 6 = \dfrac{9 + 24}{4} = \dfrac{33}{4}.$$

Hence $$x = \dfrac{3}{4}$$ and $$y = \dfrac{33}{4}.$$

Apply the identity $$(a - b)^3 = a^3 - 3a^2 b + 3a b^2 - b^3$$ with $$a = 5$$ and $$b = 3i$$:

$$(5 - 3i)^3 = 5^3 - 3(5)^2(3i) + 3(5)(3i)^2 - (3i)^3.$$

Compute each term, using $$i^2 = -1$$ and $$i^3 = -i$$:

$$5^3 = 125,$$
$$3(5)^2(3i) = 3 \cdot 25 \cdot 3i = 225 i,$$
$$3(5)(3i)^2 = 15 \cdot 9 i^2 = 135 \cdot (-1) = -135,$$
$$(3i)^3 = 27 i^3 = 27 \cdot (-i) = -27 i.$$

Substituting back:

$$(5 - 3i)^3 = 125 - 225 i + (-135) - (-27 i)$$
$$= 125 - 135 + (-225 + 27)\,i$$
$$= -10 - 198\,i.$$

First write $$\sqrt{-2}$$ in the standard form $$\sqrt{-2} = \sqrt{2}\,i$$. Then

$$\left(-\sqrt{3} + \sqrt{-2}\right)\left(2\sqrt{3} - i\right) = \left(-\sqrt{3} + \sqrt{2}\,i\right)\left(2\sqrt{3} - i\right).$$

Expand by distributing each term:

$$= (-\sqrt{3})(2\sqrt{3}) + (-\sqrt{3})(-i) + (\sqrt{2}\,i)(2\sqrt{3}) + (\sqrt{2}\,i)(-i)$$

$$= -2 \cdot 3 + \sqrt{3}\,i + 2\sqrt{6}\,i - \sqrt{2}\,i^2.$$

Using $$i^2 = -1$$:

$$= -6 + \sqrt{3}\,i + 2\sqrt{6}\,i + \sqrt{2}$$

$$= \left(\sqrt{2} - 6\right) + \left(\sqrt{3} + 2\sqrt{6}\right) i.$$

Let $$z = 2 - 3i$$. The multiplicative inverse is $$z^{-1} = \dfrac{1}{z} = \dfrac{\bar{z}}{|z|^2}.$$

The conjugate is $$\bar{z} = 2 + 3i$$, and the modulus squared is

$$|z|^2 = 2^2 + (-3)^2 = 4 + 9 = 13.$$

Hence

$$z^{-1} = \dfrac{2 + 3i}{13} = \dfrac{2}{13} + \dfrac{3}{13}\, i.$$

Check: $$(2 - 3i)\left(\dfrac{2 + 3i}{13}\right) = \dfrac{4 + 9}{13} = 1.$$

Multiply the real coefficients together and the $$i$$-factors together, using $$i^2 = -1$$:

$$(5i)\left(-\dfrac{3}{5} i\right) = 5 \cdot \left(-\dfrac{3}{5}\right) \cdot i^2 = -3 \cdot (-1) = 3.$$

In the form $$a + ib$$, this is $$3 + 0\,i.$$

Reduce each exponent modulo $$4$$ (since $$i^4 = 1$$):

$$i^9 = i^{4 \cdot 2 + 1} = (i^4)^2 \cdot i = 1 \cdot i = i.$$

$$i^{19} = i^{4 \cdot 4 + 3} = (i^4)^4 \cdot i^3 = 1 \cdot (-i) = -i.$$

Therefore

$$i^9 + i^{19} = i + (-i) = 0.$$

In the form $$a + ib$$, this is $$0 + 0\,i.$$

Write $$i^{-39} = \dfrac{1}{i^{39}}$$ and reduce $$i^{39}$$ using $$i^4 = 1$$:

$$i^{39} = i^{4 \cdot 9 + 3} = (i^4)^9 \cdot i^3 = 1 \cdot (-i) = -i.$$

Therefore

$$i^{-39} = \dfrac{1}{-i} = \dfrac{1}{-i} \cdot \dfrac{i}{i} = \dfrac{i}{-i^2} = \dfrac{i}{1} = i.$$

In the form $$a + ib$$, this is $$0 + i.$$

Distribute each multiplier:

$$3(7 + 7i) = 21 + 21 i.$$

$$i(7 + 7i) = 7 i + 7 i^2 = 7 i - 7 = -7 + 7 i.$$

Add the two results:

$$(21 + 21 i) + (-7 + 7 i) = (21 - 7) + (21 + 7) i = 14 + 28 i.$$

Remove the brackets and combine real and imaginary parts separately:

$$(1 - i) - (-1 + 6 i) = 1 - i + 1 - 6 i.$$

$$= (1 + 1) + (-1 - 6) i = 2 - 7 i.$$

Subtract real parts and imaginary parts separately:

$$\text{Real part: } \dfrac{1}{5} - 4 = \dfrac{1 - 20}{5} = -\dfrac{19}{5}.$$

$$\text{Imaginary part: } \dfrac{2}{5} - \dfrac{5}{2} = \dfrac{4 - 25}{10} = -\dfrac{21}{10}.$$

Therefore

$$\left(\dfrac{1}{5} + \dfrac{2}{5}\,i\right) - \left(4 + \dfrac{5}{2}\,i\right) = -\dfrac{19}{5} - \dfrac{21}{10}\,i.$$

First add the two complex numbers inside the bracket:

$$\left(\dfrac{1}{3} + \dfrac{7}{3}\,i\right) + \left(4 + \dfrac{1}{3}\,i\right) = \left(\dfrac{1}{3} + 4\right) + \left(\dfrac{7}{3} + \dfrac{1}{3}\right) i = \dfrac{13}{3} + \dfrac{8}{3}\,i.$$

Now subtract $$\left(-\dfrac{4}{3} + i\right)$$ from this result:

$$\left(\dfrac{13}{3} + \dfrac{8}{3}\,i\right) - \left(-\dfrac{4}{3} + i\right) = \left(\dfrac{13}{3} + \dfrac{4}{3}\right) + \left(\dfrac{8}{3} - 1\right) i.$$

$$= \dfrac{17}{3} + \dfrac{8 - 3}{3}\, i = \dfrac{17}{3} + \dfrac{5}{3}\,i.$$

Write $$(1 - i)^4 = \left[(1 - i)^2\right]^2.$$ First square $$1 - i$$:

$$(1 - i)^2 = 1^2 - 2 \cdot 1 \cdot i + i^2 = 1 - 2 i - 1 = -2 i.$$

Now square the result:

$$(-2 i)^2 = 4 i^2 = 4 \cdot (-1) = -4.$$

In the form $$a + ib$$, this is $$-4 + 0\,i.$$

Apply $$(a + b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3$$ with $$a = \dfrac{1}{3}$$ and $$b = 3 i$$:

$$a^3 = \dfrac{1}{27}.$$

$$3 a^2 b = 3 \cdot \dfrac{1}{9} \cdot 3 i = i.$$

$$3 a b^2 = 3 \cdot \dfrac{1}{3} \cdot (3 i)^2 = 1 \cdot 9 i^2 = -9.$$

$$b^3 = (3 i)^3 = 27 i^3 = -27 i.$$

Adding everything:

$$\left(\dfrac{1}{3} + 3 i\right)^3 = \dfrac{1}{27} - 9 + i - 27 i = \left(\dfrac{1}{27} - \dfrac{243}{27}\right) + (1 - 27) i = -\dfrac{242}{27} - 26\, i.$$

Apply $$(a + b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3$$ with $$a = -2$$ and $$b = -\dfrac{1}{3} i$$:

$$a^3 = (-2)^3 = -8.$$

$$3 a^2 b = 3 \cdot 4 \cdot \left(-\dfrac{1}{3}\,i\right) = -4 i.$$

$$3 a b^2 = 3 \cdot (-2) \cdot \left(-\dfrac{1}{3}\,i\right)^2 = -6 \cdot \dfrac{i^2}{9} = -6 \cdot \dfrac{-1}{9} = \dfrac{2}{3}.$$

$$b^3 = \left(-\dfrac{1}{3}\,i\right)^3 = -\dfrac{1}{27}\, i^3 = -\dfrac{1}{27} \cdot (-i) = \dfrac{1}{27}\, i.$$

Adding all four terms:

$$\left(-2 - \dfrac{1}{3}\,i\right)^3 = -8 + \dfrac{2}{3} + \left(-4 + \dfrac{1}{27}\right) i.$$

Simplify the real and imaginary parts:

$$-8 + \dfrac{2}{3} = -\dfrac{24}{3} + \dfrac{2}{3} = -\dfrac{22}{3},$$

$$-4 + \dfrac{1}{27} = -\dfrac{108}{27} + \dfrac{1}{27} = -\dfrac{107}{27}.$$

Hence $$\left(-2 - \dfrac{1}{3}\,i\right)^3 = -\dfrac{22}{3} - \dfrac{107}{27}\,i.$$

For a complex number $$z = a + ib$$, its multiplicative inverse is $$z^{-1} = \dfrac{\bar{z}}{|z|^2} = \dfrac{a - ib}{a^2 + b^2}.$$

Here $$z = 4 - 3 i$$, so $$\bar{z} = 4 + 3 i$$ and $$|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25.$$

Therefore

$$z^{-1} = \dfrac{4 + 3 i}{25} = \dfrac{4}{25} + \dfrac{3}{25}\,i.$$

Using $$z^{-1} = \dfrac{\bar{z}}{|z|^2}$$ with $$z = \sqrt{5} + 3 i$$:

$$\bar{z} = \sqrt{5} - 3 i, \qquad |z|^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14.$$

Hence

$$z^{-1} = \dfrac{\sqrt{5} - 3 i}{14} = \dfrac{\sqrt{5}}{14} - \dfrac{3}{14}\,i.$$

Here $$z = -i = 0 + (-1) i$$. Its conjugate is $$\bar{z} = i$$ and $$|z|^2 = 0^2 + (-1)^2 = 1.$$

Therefore

$$z^{-1} = \dfrac{\bar{z}}{|z|^2} = \dfrac{i}{1} = i.$$

Check: $$(-i)(i) = -i^2 = 1.$$ In the form $$a + ib$$, the inverse is $$0 + i.$$

Simplify the numerator using $$(a + b)(a - b) = a^2 - b^2$$:

$$(3 + i\sqrt{5})(3 - i\sqrt{5}) = 3^2 - (i\sqrt{5})^2 = 9 - i^2 \cdot 5 = 9 - (-5) = 14.$$

Simplify the denominator:

$$(\sqrt{3} + \sqrt{2}\,i) - (\sqrt{3} - \sqrt{2}\,i) = \sqrt{3} + \sqrt{2}\,i - \sqrt{3} + \sqrt{2}\,i = 2\sqrt{2}\,i.$$

Therefore the expression equals

$$\dfrac{14}{2\sqrt{2}\,i} = \dfrac{7}{\sqrt{2}\,i}.$$

Rationalise by multiplying numerator and denominator by $$-i/\sqrt{2}$$ (or just by $$i$$, then simplify):

$$\dfrac{7}{\sqrt{2}\,i} \cdot \dfrac{i}{i} = \dfrac{7 i}{\sqrt{2}\, i^2} = \dfrac{7 i}{-\sqrt{2}} = -\dfrac{7}{\sqrt{2}}\,i = -\dfrac{7\sqrt{2}}{2}\,i.$$

In the form $$a + ib$$, this is $$0 - \dfrac{7\sqrt{2}}{2}\,i.$$

First express the given quotient in the standard form $$a + ib$$; its conjugate will then be $$a - ib$$.

Numerator:

$$(3 - 2i)(2 + 3i) = 6 + 9 i - 4 i - 6 i^2 = 6 + 5 i + 6 = 12 + 5 i.$$

Denominator:

$$(1 + 2i)(2 - i) = 2 - i + 4 i - 2 i^2 = 2 + 3 i + 2 = 4 + 3 i.$$

Hence the quotient is

$$\dfrac{12 + 5 i}{4 + 3 i}.$$

Multiply numerator and denominator by the conjugate $$4 - 3 i$$ of the denominator:

$$\dfrac{(12 + 5 i)(4 - 3 i)}{(4 + 3 i)(4 - 3 i)} = \dfrac{48 - 36 i + 20 i - 15 i^2}{16 + 9} = \dfrac{48 - 16 i + 15}{25} = \dfrac{63 - 16 i}{25}.$$

So the expression equals $$\dfrac{63}{25} - \dfrac{16}{25}\,i.$$ Its conjugate is

$$\dfrac{63}{25} + \dfrac{16}{25}\,i.$$

Multiply the numerator and denominator on the right-hand side by $$a + ib$$ (the conjugate of $$a - ib$$):

$$x + iy = \dfrac{a + ib}{a - ib} \cdot \dfrac{a + ib}{a + ib} = \dfrac{(a + ib)^2}{(a - ib)(a + ib)} = \dfrac{(a + ib)^2}{a^2 + b^2}.$$

Expand the numerator:

$$(a + ib)^2 = a^2 + 2 a b\,i + (ib)^2 = a^2 - b^2 + 2 a b\,i.$$

Therefore

$$x + iy = \dfrac{a^2 - b^2}{a^2 + b^2} + \dfrac{2 a b}{a^2 + b^2}\, i.$$

Comparing real and imaginary parts gives

$$x = \dfrac{a^2 - b^2}{a^2 + b^2}, \qquad y = \dfrac{2 a b}{a^2 + b^2}.$$

Now compute $$x^2 + y^2$$:

$$x^2 + y^2 = \dfrac{(a^2 - b^2)^2 + (2 a b)^2}{(a^2 + b^2)^2} = \dfrac{a^4 - 2 a^2 b^2 + b^4 + 4 a^2 b^2}{(a^2 + b^2)^2}.$$

$$= \dfrac{a^4 + 2 a^2 b^2 + b^4}{(a^2 + b^2)^2} = \dfrac{(a^2 + b^2)^2}{(a^2 + b^2)^2} = 1.$$

Hence $$x^2 + y^2 = 1$$ as required.

Reduce each term using $$i^4 = 1$$.

Term 1:

$$i^{18} = i^{4 \cdot 4 + 2} = (i^4)^4 \cdot i^2 = 1 \cdot (-1) = -1.$$

Term 2: Note that $$\dfrac{1}{i} = \dfrac{1}{i} \cdot \dfrac{i}{i} = \dfrac{i}{i^2} = -i.$$ Hence

$$\left(\dfrac{1}{i}\right)^{25} = (-i)^{25} = (-1)^{25} \cdot i^{25} = -1 \cdot i^{4 \cdot 6 + 1} = -1 \cdot i = -i.$$

Therefore the bracket simplifies to

$$i^{18} + \left(\dfrac{1}{i}\right)^{25} = -1 + (-i) = -1 - i.$$

Now cube the result using $$(a + b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3$$ with $$a = -1, b = -i$$, or more easily $$(-1 - i)^3 = -(1 + i)^3$$:

$$(1 + i)^3 = 1 + 3 i + 3 i^2 + i^3 = 1 + 3 i - 3 - i = -2 + 2 i.$$

So

$$(-1 - i)^3 = -(1 + i)^3 = -(-2 + 2 i) = 2 - 2 i.$$

Let $$z_1 = x_1 + i\,y_1$$ and $$z_2 = x_2 + i\,y_2$$, where $$x_1, y_1, x_2, y_2$$ are real. Then

$$\mathrm{Re}\,z_1 = x_1, \quad \mathrm{Im}\,z_1 = y_1, \quad \mathrm{Re}\,z_2 = x_2, \quad \mathrm{Im}\,z_2 = y_2.$$

Compute the product:

$$z_1 z_2 = (x_1 + i\,y_1)(x_2 + i\,y_2) = x_1 x_2 + i\,x_1 y_2 + i\,y_1 x_2 + i^2\,y_1 y_2.$$

Using $$i^2 = -1$$:

$$z_1 z_2 = (x_1 x_2 - y_1 y_2) + i\,(x_1 y_2 + y_1 x_2).$$

Therefore

$$\mathrm{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = \mathrm{Re}\,z_1 \cdot \mathrm{Re}\,z_2 - \mathrm{Im}\,z_1 \cdot \mathrm{Im}\,z_2,$$

as required.

Step 1. Rationalise each fraction in the first bracket.

$$\dfrac{1}{1 - 4 i} = \dfrac{1 + 4 i}{(1 - 4 i)(1 + 4 i)} = \dfrac{1 + 4 i}{1 + 16} = \dfrac{1 + 4 i}{17}.$$

$$\dfrac{2}{1 + i} = \dfrac{2 (1 - i)}{(1 + i)(1 - i)} = \dfrac{2 (1 - i)}{2} = 1 - i.$$

Subtract:

$$\dfrac{1 + 4 i}{17} - (1 - i) = \dfrac{1 + 4 i - 17(1 - i)}{17} = \dfrac{1 + 4 i - 17 + 17 i}{17} = \dfrac{-16 + 21 i}{17}.$$

Step 2. Rationalise the second factor.

$$\dfrac{3 - 4 i}{5 + i} = \dfrac{(3 - 4 i)(5 - i)}{(5 + i)(5 - i)} = \dfrac{15 - 3 i - 20 i + 4 i^2}{25 + 1} = \dfrac{15 - 23 i - 4}{26} = \dfrac{11 - 23 i}{26}.$$

Step 3. Multiply the two results.

$$\dfrac{-16 + 21 i}{17} \cdot \dfrac{11 - 23 i}{26} = \dfrac{(-16 + 21 i)(11 - 23 i)}{17 \cdot 26} = \dfrac{(-16 + 21 i)(11 - 23 i)}{442}.$$

Expand the numerator:

$$(-16)(11) + (-16)(-23 i) + (21 i)(11) + (21 i)(-23 i)$$
$$= -176 + 368 i + 231 i - 483 i^2 = -176 + 599 i + 483 = 307 + 599 i.$$

Therefore

$$\left(\dfrac{1}{1 - 4 i} - \dfrac{2}{1 + i}\right)\left(\dfrac{3 - 4 i}{5 + i}\right) = \dfrac{307 + 599 i}{442} = \dfrac{307}{442} + \dfrac{599}{442}\,i.$$

Square both sides of the given equation:

$$(x - i y)^2 = \dfrac{a - i b}{c - i d}. \qquad (1)$$

Take the complex conjugate of both sides of (1). The conjugate of a quotient equals the quotient of the conjugates, and the conjugate of a square equals the square of the conjugate, so

$$(x + i y)^2 = \dfrac{a + i b}{c + i d}. \qquad (2)$$

Multiply equations (1) and (2):

$$(x - i y)^2 (x + i y)^2 = \dfrac{(a - i b)(a + i b)}{(c - i d)(c + i d)}.$$

The left side equals $$\left[(x - i y)(x + i y)\right]^2 = (x^2 + y^2)^2.$$

The right side equals $$\dfrac{a^2 + b^2}{c^2 + d^2}$$ (using $$(p - i q)(p + i q) = p^2 + q^2$$).

Hence

$$(x^2 + y^2)^2 = \dfrac{a^2 + b^2}{c^2 + d^2},$$

as required.

Compute the numerator and denominator separately.

Numerator:

$$z_1 + z_2 + 1 = (2 - i) + (1 + i) + 1 = 4 + 0\,i = 4.$$

Denominator:

$$z_1 - z_2 + 1 = (2 - i) - (1 + i) + 1 = 2 - i - 1 - i + 1 = 2 - 2 i.$$

Therefore

$$\left|\dfrac{z_1 + z_2 + 1}{z_1 - z_2 + 1}\right| = \left|\dfrac{4}{2 - 2 i}\right| = \dfrac{|4|}{|2 - 2 i|} = \dfrac{4}{\sqrt{2^2 + (-2)^2}} = \dfrac{4}{\sqrt{8}} = \dfrac{4}{2\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}.$$

Expand $$(x + i)^2$$:

$$(x + i)^2 = x^2 + 2 x\,i + i^2 = (x^2 - 1) + 2 x\,i.$$

Therefore

$$a + i b = \dfrac{(x^2 - 1) + 2 x\,i}{2 x^2 + 1} = \dfrac{x^2 - 1}{2 x^2 + 1} + \dfrac{2 x}{2 x^2 + 1}\,i.$$

Comparing real and imaginary parts:

$$a = \dfrac{x^2 - 1}{2 x^2 + 1}, \qquad b = \dfrac{2 x}{2 x^2 + 1}.$$

Now compute $$a^2 + b^2$$:

$$a^2 + b^2 = \dfrac{(x^2 - 1)^2 + (2 x)^2}{(2 x^2 + 1)^2} = \dfrac{(x^2 - 1)^2 + 4 x^2}{(2 x^2 + 1)^2}.$$

Simplify the numerator:

$$(x^2 - 1)^2 + 4 x^2 = x^4 - 2 x^2 + 1 + 4 x^2 = x^4 + 2 x^2 + 1 = (x^2 + 1)^2.$$

Hence

$$a^2 + b^2 = \dfrac{(x^2 + 1)^2}{(2 x^2 + 1)^2},$$

as required.

The conjugate of $$-6 - 24 i$$ is $$-6 + 24 i.$$ So we need

$$(x - i y)(3 + 5 i) = -6 + 24 i.$$

Expand the left-hand side:

$$(x - i y)(3 + 5 i) = 3 x + 5 x\,i - 3 y\,i - 5 y\,i^2 = 3 x + 5 x\,i - 3 y\,i + 5 y = (3 x + 5 y) + (5 x - 3 y)\,i.$$

Equate real and imaginary parts with $$-6 + 24 i$$:

$$3 x + 5 y = -6, \qquad 5 x - 3 y = 24.$$

Solve this linear system. Multiply the first by $$3$$ and the second by $$5$$:

$$9 x + 15 y = -18, \qquad 25 x - 15 y = 120.$$

Add the equations:

$$34 x = 102 \implies x = 3.$$

Substitute back into $$3 x + 5 y = -6$$:

$$3(3) + 5 y = -6 \implies 5 y = -15 \implies y = -3.$$

Hence $$x = 3$$ and $$y = -3.$$

Combine the two fractions over the common denominator $$(1 - i)(1 + i) = 1 - i^2 = 2$$:

$$\dfrac{1 + i}{1 - i} - \dfrac{1 - i}{1 + i} = \dfrac{(1 + i)^2 - (1 - i)^2}{(1 - i)(1 + i)}.$$

Expand each square:

$$(1 + i)^2 = 1 + 2 i + i^2 = 2 i, \qquad (1 - i)^2 = 1 - 2 i + i^2 = -2 i.$$

So the numerator is $$2 i - (-2 i) = 4 i,$$ and the denominator is $$2.$$ Hence

$$\dfrac{1 + i}{1 - i} - \dfrac{1 - i}{1 + i} = \dfrac{4 i}{2} = 2 i.$$

The modulus is

$$|2 i| = \sqrt{0^2 + 2^2} = 2.$$

Expand $$(x + i y)^3$$ using the binomial expansion, with $$i^2 = -1$$ and $$i^3 = -i$$:

$$(x + i y)^3 = x^3 + 3 x^2 (i y) + 3 x (i y)^2 + (i y)^3$$
$$= x^3 + 3 x^2 y\,i - 3 x y^2 - y^3\,i$$
$$= (x^3 - 3 x y^2) + (3 x^2 y - y^3)\,i.$$

Comparing with $$(x + i y)^3 = u + i v$$:

$$u = x^3 - 3 x y^2 = x (x^2 - 3 y^2), \qquad v = 3 x^2 y - y^3 = y (3 x^2 - y^2).$$

Therefore (assuming $$x \neq 0$$ and $$y \neq 0$$)

$$\dfrac{u}{x} = x^2 - 3 y^2, \qquad \dfrac{v}{y} = 3 x^2 - y^2.$$

Adding:

$$\dfrac{u}{x} + \dfrac{v}{y} = (x^2 - 3 y^2) + (3 x^2 - y^2) = 4 x^2 - 4 y^2 = 4 (x^2 - y^2),$$

as required.

Recall two facts:

• $$|\bar{z}| = |z|$$ for every complex number $$z$$.
• $$z \bar{z} = |z|^2$$, so $$|\beta| = 1 \implies \beta \bar{\beta} = 1.$$

Multiply the modulus of the denominator by $$|\bar{\beta}| = |\beta| = 1$$ (which does not change its value):

$$|1 - \bar{\alpha}\beta| = |\bar{\beta}| \cdot |1 - \bar{\alpha}\beta| = \left|\bar{\beta}(1 - \bar{\alpha}\beta)\right| = \left|\bar{\beta} - \bar{\alpha}\,\beta \bar{\beta}\right| = |\bar{\beta} - \bar{\alpha}|.$$

(In the last step we used $$\beta \bar{\beta} = 1$$.) Now

$$|\bar{\beta} - \bar{\alpha}| = |\overline{\beta - \alpha}| = |\beta - \alpha|.$$

Therefore

$$\left|\dfrac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right| = \dfrac{|\beta - \alpha|}{|1 - \bar{\alpha}\beta|} = \dfrac{|\beta - \alpha|}{|\beta - \alpha|} = 1.$$

(The denominator does not vanish because $$\alpha \neq \beta$$, so $$|\beta - \alpha| > 0.$$)

Compute the modulus:

$$|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}.$$

So the equation becomes

$$(\sqrt{2})^x = 2^x \iff 2^{x/2} = 2^x.$$

Since the exponential function $$t \mapsto 2^t$$ is one-to-one, the exponents must be equal:

$$\dfrac{x}{2} = x \iff x - \dfrac{x}{2} = 0 \iff \dfrac{x}{2} = 0 \iff x = 0.$$

The only solution is $$x = 0$$, which is excluded by the requirement that $$x$$ be a non-zero integer. Hence the number of non-zero integral solutions is $$0.$$

Use the multiplicative property of the modulus: for any complex numbers $$z_1, z_2$$, $$|z_1 z_2| = |z_1|\,|z_2|.$$ This extends to any finite product.

Take the modulus of both sides of the given equation:

$$|(a + i b)(c + i d)(e + i f)(g + i h)| = |A + i B|.$$

By the multiplicative property,

$$|a + i b|\,|c + i d|\,|e + i f|\,|g + i h| = |A + i B|.$$

Now square both sides; recall $$|p + i q|^2 = p^2 + q^2$$, so

$$|a + i b|^2 = a^2 + b^2, \ |c + i d|^2 = c^2 + d^2, \ |e + i f|^2 = e^2 + f^2, \ |g + i h|^2 = g^2 + h^2,$$
$$|A + i B|^2 = A^2 + B^2.$$

Therefore

$$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2,$$

which is the required identity.

First simplify $$\dfrac{1 + i}{1 - i}$$ by multiplying numerator and denominator by the conjugate $$1 + i$$ of the denominator:

$$\dfrac{1 + i}{1 - i} = \dfrac{(1 + i)^2}{(1 - i)(1 + i)} = \dfrac{1 + 2 i + i^2}{1 - i^2} = \dfrac{2 i}{2} = i.$$

So the equation becomes

$$i^m = 1.$$

The powers of $$i$$ cycle as $$i, -1, -i, 1$$ with period $$4$$; specifically, $$i^m = 1$$ iff $$m$$ is a multiple of $$4.$$ The least positive integer that satisfies this is

$$m = 4.$$