Chapter 4 — Complex Numbers and Quadratic Equations

We evaluate \[ \left[ i^{18}+\left( \dfrac{1}{i}\right)^{25}\right]^3 \]

Reduce powers: \[ i^{18}=i^{(18 \bmod 4)}=i^2=-1 \] \[ \left(\frac{1}{i}\right)^{25}=i^{-25}=i^{(-25 \bmod 4)}=i^{-1}=-i \]

Therefore, \[ i^{18}+\left(\frac{1}{i}\right)^{25}=-1-i \]

Now cube: \[ (-1-i)^2 = 1 + 2i + i^2 = 2i \] \[ (-1-i)^3 = 2i(-1-i)= -2i -2i^2 = 2 - 2i \]

Final Answer: \(2-2i\)

Let \[ z_1 = x_1 + iy_1,\quad z_2 = x_2 + iy_2 \]

Multiply: \[ \begin{aligned} z_1 z_2 &= (x_1 + iy_1)(x_2 + iy_2) \\ &= x_1 x_2 + i x_1 y_2 + i y_1 x_2 + i^2 y_1 y_2 \end{aligned} \]

Using \(i^2 = -1\), \[ z_1 z_2 = x_1 x_2 - y_1 y_2 + i(x_1 y_2 + y_1 x_2) \]

Hence, the real part is \[ \Re(z_1 z_2) = x_1 x_2 - y_1 y_2 \]

Substituting \[ x_1=\Re z_1,\quad x_2=\Re z_2,\quad y_1=\Im z_1,\quad y_2=\Im z_2 \]

\[ \Re(z_1 z_2) = \Re z_1 \Re z_2 - \Im z_1 \Im z_2 \]

Hence proved.

Start with: \[ \left( \dfrac{1}{1-4i}-\dfrac{2}{1+i}\right)\left( \dfrac{3-4i}{5+i}\right) \]

Combine first bracket: \[ \dfrac{1}{1-4i}-\dfrac{2}{1+i} = \dfrac{(1+i)-2(1-4i)}{(1-4i)(1+i)} \]

Simplify numerator: \[ (1+i)-2(1-4i) = 1+i-2+8i = -1+9i \]

So expression becomes: \[ \dfrac{-1+9i}{(1-4i)(1+i)} \cdot \dfrac{3-4i}{5+i} = \dfrac{(-1+9i)(3-4i)}{(1-4i)(1+i)(5+i)} \]

Numerator: \[ \begin{align} (-1+9i)(3-4i) &= -3+4i+27i-36i^2\\ &= -3+31i+36\\ &= 33+31i \end{align} \]

Denominator: \[ (1-4i)(1+i)=1+i-4i-4i^2=5-3i \] \[ (5-3i)(5+i)=25+5i-15i-3i^2=28-10i \]

Thus, \[ \frac{33+31i}{28-10i} \]

Rationalize: \[ \frac{33+31i}{28-10i}\cdot\frac{28+10i}{28+10i} = \frac{(33+31i)(28+10i)}{28^2+10^2} \]

Numerator: \[ \begin{aligned} &= 924+330i+868i+310i^2\\ &= 924+1198i-310\\ &= 614+1198i \end{aligned} \]

Denominator: \[ 28^2+10^2 = 784+100 = 884 \]

\[ \begin{aligned} &= \frac{614}{884} + \frac{1198}{884}i\\ &= \frac{307}{442} + \frac{599}{442}i \end{aligned} \]

Given: \[ x-iy=\sqrt{\dfrac{a-ib}{c-id}} \]

Taking conjugate: \[ x+iy=\sqrt{\dfrac{a+ib}{c+id}} \]

Multiply: \[ (x-iy)(x+iy) = \sqrt{\dfrac{a-ib}{c-id}\cdot\dfrac{a+ib}{c+id}} \]

LHS: \[ (x-iy)(x+iy)=x^2 + y^2 \]

RHS: \[ \begin{aligned} \dfrac{a-ib}{c-id}\cdot\dfrac{a+ib}{c+id} &= \dfrac{(a-ib)(a+ib)}{(c-id)(c+id)}\\ &= \dfrac{a^2+b^2}{c^2+d^2} \end{aligned} \]

Therefore, \[ x^2+y^2 = \sqrt{\dfrac{a^2+b^2}{c^2+d^2}} \]

Squaring both sides: \[ (x^2+y^2)^2 = \dfrac{a^2+b^2}{c^2+d^2} \]

Hence proved.

Given: \[ \begin{aligned} z_1 &= 2 - i,\\ z_2 &= 1 + i \end{aligned" \]

Numerator: \[ \begin{aligned z_1 + z_2 + 1 &= (2 - i) + (1 + i) + 1\\ &= 4 \end{aligned" \]

Denominator: \[ \begin{aligned} z_1 - z_2 + 1 &= (2 - i) - (1 + i) + 1\\ &= 2 - 2i \end{aligned} \]

Taking moduli: \[ \begin{aligned} |4| &= 4\\ |2 - 2i| &= \sqrt{2^2 + (-2)^2} \\&= \sqrt{8} \end{aligned} \]

Using modulus property: \[ \begin{aligned} \left|\frac{z_1+z_2+1}{z_1-z_2+1}\right| &= \frac{4}{\sqrt{8}}\\ &= \sqrt{2} \end{aligned} \]

Final Answer: \(\sqrt{2}\)

Given: \[ a+ib=\dfrac{(x+i)^2}{2x^2+1} \]

Taking conjugate: \[ a-ib=\dfrac{(x-i)^2}{2x^2+1} \]

Multiply: \[ (a+ib)(a-ib) = \dfrac{(x+i)^2 (x-i)^2}{(2x^2+1)^2} \]

LHS: \[ a^2 + b^2 \]

RHS: \[ (x+i)(x-i)=x^2+1 \] \[ \Rightarrow (x+i)^2 (x-i)^2 = (x^2+1)^2 \]

\[ a^2 + b^2 = \dfrac{(x^2+1)^2}{(2x^2+1)^2} \]

Hence proved.

Given: \[ z_1 = 2 - i,\quad z_2 = -2 + i \]

Step 1: Compute product \[ \begin{aligned} z_1 z_2 &= (2 - i)(-2 + i)\\ &= -4 + 2i + 2i - i^2\\ &= -3 + 4i \end{aligned} \]

Step 2: Conjugate \[ \overline{z_1} = 2 + i \]

Step 3: Divide \[ \frac{z_1 z_2}{\overline{z_1}} = \frac{-3 + 4i}{2 + i} \]

Rationalize: \[ = \frac{(-3 + 4i)(2 - i)}{(2+i)(2-i)} \]

Numerator: \[ \begin{aligned} &= -6 + 3i + 8i - 4i^2\\ &= -6 + 11i + 4\\ &= -2 + 11i \end{aligned} \]

Denominator: \[ = 4 - i^2 = 5 \]

\[ = \frac{-2 + 11i}{5} \]

\[ \Re\left(\frac{z_1 z_2}{\overline{z_1}}\right) = -\frac{2}{5} \]

Step 4: \[ z_1 \overline{z_1} = (2 - i)(2 + i) = 5 \]

\[ \frac{1}{z_1 \overline{z_1}} = \frac{1}{5} \]

\[ \Im\left(\frac{1}{z_1 \overline{z_1}}\right) = 0 \]

Given: \[ (x - iy)(3 + 5i) = \text{conjugate of }(-6 - 24i) \]

Conjugate: \[ -6 - 24i \Rightarrow -6 + 24i \]

So, \[ (x - iy)(3 + 5i) = -6 + 24i \]

Expand: \[ (x - iy)(3 + 5i) = 3x + 5iy - 3iy - 5i^2 y \] \[ = 3x + 5y + i(5x - 3y) \]

Compare with \(-6 + 24i\):

Real part: \[ 3x + 5y = -6 \tag{1} \]

Imaginary part: \[ 5x - 3y = 24 \tag{2} \]

Solve: Multiply (2) by 5: \[ 25x - 15y = 120 \]

Multiply (1) by 3: \[ 9x + 15y = -18 \]

Add: \[ \begin{aligned} 34x &= 102 \\ \Rightarrow x &= 3 \end{aligned} \]

Substitute in (2): \[ \begin{aligned} 5(3) - 3y &= 24 \Rightarrow 15 - 3y \\&= 24\\ \Rightarrow y &= -3 \end{aligned} \]

Final Answer: \(x = 3,\; y = -3\)

Consider: \[ \frac{1+i}{1-i}-\frac{1-i}{1+i} \]

Take common denominator: \[ = \frac{(1+i)^2 - (1-i)^2}{(1-i)(1+i)} \]

Expand numerator: \[ \begin{aligned} (1+i)^2 &= 1 + 2i + i^2 &\\= 2i \end{aligned} \] \[ \begin{aligned} (1-i)^2 &= 1 - 2i + i^2 &\\= -2i \end{aligned} \]

So numerator: \[ 2i - (-2i) = 4i \]

Denominator: \[ \begin{aligned} (1-i)(1+i) &= 1 - i^2 \\&= 2 \end{aligned} \]

Hence: \[ = \frac{4i}{2} = 2i \]

Modulus: \[ \begin{aligned} |2i| &= \sqrt{0^2 + 2^2} \\&= 2 \end{aligned} \]

Final Answer: \(2\)

Given: \[ (x+iy)^3 = u + iv \]

Expand: \[ (x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 \]

Using \(i^2=-1,\; i^3=-i\): \[ = x^3 + 3ix^2y + 3x(-y^2) + (-i)y^3 \]

\[ = (x^3 - 3xy^2) + i(3x^2y - y^3) \]

Hence, \[ \begin{aligned} u &= x^3 - 3xy^2,\\ v &= 3x^2y - y^3 \end{aligned} \]

Compute: \[ \begin{aligned} \frac{u}{x} &= \frac{x^3 - 3xy^2}{x} \\&= x^2 - 3y^2 \end{aligned} \]

\[ \begin{aligned} \frac{v}{y} &= \frac{3x^2y - y^3}{y} \\&= 3x^2 - y^2 \end{aligned} \]

Add: \[ \frac{u}{x} + \frac{v}{y} = (x^2 - 3y^2) + (3x^2 - y^2) \]

\[ \begin{aligned] &= 4x^2 - 4y^2 \\&= 4(x^2 - y^2) \END{alignED} \]

Hence proved.

Consider: \[ \left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|^2 = \frac{(\beta-\alpha)(\overline{\beta}-\overline{\alpha})} {(1-\overline{\alpha}\beta)(1-\alpha\overline{\beta})} \]

Expand numerator: \[ = |\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2 \]

Expand denominator: \[ = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2|\beta|^2 \]

Given: \[ \begin{aligned} |\beta| &= 1 \\\Rightarrow |\beta|^2 &= 1 \end{aligned} \]

Substitute: \[ \frac{1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2} {1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2} \]

Numerator and denominator are identical:

\[ \begin{aligned} \left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|^2 &= 1\\ \Rightarrow \left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right| &= 1 \end{aligned} \]

First compute modulus: \[ \begin{aligned} |1-i| &= \sqrt{1^2 + (-1)^2} \\&= \sqrt{2} \end{aligned} \]

Given equation: \[ \begin{aligned} |1-i|^x &= 2^x \Rightarrow (\sqrt{2})^x \\&= 2^x \end{aligned} \]

Write in same base: \[ \begin{aligned} (2^{1/2})^x &= 2^x\\ \Rightarrow 2^{x/2} &= 2^x \end{aligned} \]

Equate exponents: \[ \begin{aligned} \frac{x}{2} &= x\\ \Rightarrow x &= 0 \end{aligned} \]

But required: non-zero integers

Therefore, no non-zero integer satisfies the equation

Number of solutions = 0

Given: \[ (a+ib)(c+id)(e+if)(g+ih) = A + iB \]

Take modulus: \[ |(a+ib)(c+id)(e+if)(g+ih)| = |A+iB| \]

Using property: \[ |z_1 z_2 z_3 z_4| = |z_1||z_2||z_3||z_4| \]

\[ |a+ib|\;|c+id|\;|e+if|\;|g+ih| = \sqrt{A^2 + B^2} \]

Replace each modulus: \[ \sqrt{a^2+b^2}\;\sqrt{c^2+d^2}\;\sqrt{e^2+f^2}\;\sqrt{g^2+h^2} = \sqrt{A^2+B^2} \]

Squaring both sides:

\[ (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2 \]

Hence proved.

Simplify: \[ \begin{aligned} \frac{1+i}{1-i} &= \frac{1+i}{1-i} \cdot \frac{1+i}{1+i}\\ &= \frac{(1+i)^2}{1 - i^2} \end{aligned} \]

\[ \begin{aligned} (1+i)^2 &= 1 + 2i + i^2 \\&= 2i \end{aligned} \] \[ 1 - i^2 = 2 \]

\[ \begin{aligned} \frac{1+i}{1-i} &= \frac{2i}{2} \\&= i \end{aligned} \]

Given: \[ \begin{aligned} \left(\frac{1+i}{1-i}\right)^m &= 1\\ \Rightarrow i^m &= 1 \end{aligned} \]

Using cyclic pattern: \[ i^4 = 1 \]

Least positive integer: \[ m = 4 \]