Ejercicios de la Universidad de Antioquia sobre números complejos, ecuaciones y propiedades de conjugados.

Steps to Solve

1. Solve A.1.a: $\frac{(5 + i) - (-1+7i)}{(2 - 3i) + (6 +5i)}$ Simplify the numerator and denominator separately $$ \frac{(5 + i) - (-1+7i)}{(2 - 3i) + (6 +5i)} = \frac{5 + i + 1 - 7i}{2 - 3i + 6 + 5i} = \frac{6 - 6i}{8 + 2i} $$ Multiply the numerator and denominator by the conjugate of the denominator: $$ \frac{6 - 6i}{8 + 2i} \cdot \frac{8 - 2i}{8 - 2i} = \frac{(6 - 6i)(8 - 2i)}{(8 + 2i)(8 - 2i)} = \frac{48 - 12i - 48i + 12i^2}{64 - 16i + 16i - 4i^2} $$ Since $i^2 = -1$: $$ \frac{48 - 60i - 12}{64 + 4} = \frac{36 - 60i}{68} = \frac{36}{68} - \frac{60}{68}i = \frac{9}{17} - \frac{15}{17}i $$ The complex number is $\frac{9}{17} - \frac{15}{17}i$ The conjugate of the complex number is $\frac{9}{17} + \frac{15}{17}i$ Calculate the product with its conjugate: $$ (\frac{9}{17} - \frac{15}{17}i)(\frac{9}{17} + \frac{15}{17}i) = (\frac{9}{17})^2 - (\frac{15}{17}i)^2 = \frac{81}{289} + \frac{225}{289} = \frac{306}{289} = \frac{18}{17} $$

2. Solve A.1.b: $\frac{(1-2i)^{3}}{5i^{6}}$ Expand the numerator: $$ (1 - 2i)^3 = (1 - 2i)(1 - 2i)(1 - 2i) = (1 - 4i + 4i^2)(1 - 2i) = (1 - 4i - 4)(1 - 2i) = (-3 - 4i)(1 - 2i) = -3 + 6i - 4i + 8i^2 = -3 + 2i - 8 = -11 + 2i $$ Simplify the denominator: $$ 5i^6 = 5(i^2)^3 = 5(-1)^3 = -5 $$ Then divide: $$ \frac{-11 + 2i}{-5} = \frac{11}{5} - \frac{2}{5}i $$ The complex number is $\frac{11}{5} - \frac{2}{5}i$ The conjugate of the complex number is $\frac{11}{5} + \frac{2}{5}i$ Calculate the product with its conjugate: $$ (\frac{11}{5} - \frac{2}{5}i)(\frac{11}{5} + \frac{2}{5}i) = (\frac{11}{5})^2 - (\frac{2}{5}i)^2 = \frac{121}{25} + \frac{4}{25} = \frac{125}{25} = 5 $$

3. Solve A.1.c: $[\frac{4+\sqrt{-81}}{5-\sqrt{-100}}]\cdot[\frac{\sqrt{-16}\sqrt{-25}}{\sqrt{-9}}]$ Simplify square roots: $$ \sqrt{-81}=9i, \sqrt{-100}=10i, \sqrt{-16}=4i, \sqrt{-25}=5i, \sqrt{-9}=3i $$ Substitute back into expression: $$ \frac{4+9i}{5-10i} \cdot \frac{(4i)(5i)}{3i} = \frac{4+9i}{5-10i} \cdot \frac{20i^2}{3i} = \frac{4+9i}{5-10i} \cdot \frac{-20}{3i} $$ Multiply by conjugate of $5-10i$: $$ \frac{4+9i}{5-10i} \cdot \frac{5+10i}{5+10i} = \frac{20 + 40i + 45i + 90i^2}{25 + 100} = \frac{20 + 85i - 90}{125} = \frac{-70 + 85i}{125} = \frac{-14+17i}{25} $$ Therefore, $$ \frac{-14+17i}{25} \cdot \frac{-20}{3i} = \frac{-14+17i}{25} \cdot \frac{-20}{3i} \cdot \frac{-i}{-i} = \frac{(-14+17i) \cdot 20i}{25 \cdot 3(-i^2)} = \frac{-280i + 340i^2}{75} = \frac{-340-280i}{75} = \frac{-68}{15}-\frac{56}{15}i $$ The complex number is $\frac{-68}{15}-\frac{56}{15}i$ The conjugate of the complex number is $\frac{-68}{15}+\frac{56}{15}i$ Calculate the product with its conjugate: $$ (\frac{-68}{15}-\frac{56}{15}i)(\frac{-68}{15}+\frac{56}{15}i) = (\frac{-68}{15})^2 - (\frac{56}{15}i)^2 = \frac{4624}{225} + \frac{3136}{225} = \frac{7760}{225} = \frac{1552}{45} $$

4. Solve A.1.d: $(i^{3} +2i^{13})^{4}$ Simplify the expression inside the parenthesis. $i^3 = -i$ and $i^{13} = i^{12} \cdot i = (i^4)^3 \cdot i = 1^3 \cdot i = i$, so $$ (i^3 + 2i^{13})^4 = (-i + 2i)^4 = (i)^4 = 1 $$ The complex number is $1$ The conjugate of the complex number is $1$ Calculate the product with its conjugate: $$ 1\cdot 1 = 1 $$

5. Solve A.1.e: $[(5i - 3)(2i+1)]^{-2}$ Simplify the expression inside the parenthesis. $$ (5i - 3)(2i + 1) = 10i^2 + 5i - 6i - 3 = -10 -i - 3 = -13 - i $$ Then we have $$ (-13 - i)^{-2} = \frac{1}{(-13 - i)^2} = \frac{1}{(-13 - i)(-13 - i)} = \frac{1}{169 + 13i + 13i + i^2} = \frac{1}{169 + 26i - 1} = \frac{1}{168 + 26i} $$ Multiply by the conjugate: $$ \frac{1}{168 + 26i} \cdot \frac{168 - 26i}{168 - 26i} = \frac{168 - 26i}{168^2 + 26^2} = \frac{168 - 26i}{28224 + 676} = \frac{168 - 26i}{28900} = \frac{168}{28900} - \frac{26}{28900}i = \frac{42}{7225} - \frac{13}{14450}i $$ The complex number is $\frac{42}{7225} - \frac{13}{14450}i$ The conjugate of the complex number is $\frac{42}{7225} + \frac{13}{14450}i$ Calculate the product with its conjugate: $$ (\frac{42}{7225} - \frac{13}{14450}i)(\frac{42}{7225} + \frac{13}{14450}i) = (\frac{42}{7225})^2 - (\frac{13}{14450}i)^2 = \frac{1764}{52200625} + \frac{169}{208802500} = \frac{67600+169}{208802500} = \frac{67769}{208802500} = \frac{1}{321} $$ Almost there, calculating it again $$ (\frac{42}{7225} - \frac{13}{14450}i)(\frac{42}{7225} + \frac{13}{14450}i) = (\frac{42}{7225})^2 + (\frac{13}{14450})^2 = \frac{1764}{52200625} + \frac{169}{208802500} = \frac{676 \cdot 4 + 169}{208802500} = \frac{2704+169}{208802500} = \frac{2873}{208802500} = \frac{1}{7267.3} $$ There must be some mistake here Let's go back $$ \frac{1}{168 + 26i} \cdot \frac{168 - 26i}{168 - 26i} = \frac{168 - 26i}{168^2 + 26^2} = \frac{168 - 26i}{28900} $$ The original number was $\frac{1}{(-13 - i)^2} = \frac{1}{(168 + 26i)} $ and conjugate is $\frac{1}{168 - 26i}$ Multiplying them gives: $$ \frac{1}{(168 + 26i)} \cdot \frac{1}{(168 - 26i)} = \frac{1}{168^2 + 26^2} = \frac{1}{28900} $$

6. Solve A.1.f: $[(2+6i)^{2}]^{-1}$ $$ [(2+6i)^2]^{-1} = \frac{1}{(2+6i)^2} = \frac{1}{(2+6i)(2+6i)} = \frac{1}{4 + 12i + 12i + 36i^2} = \frac{1}{4 + 24i - 36} = \frac{1}{-32 + 24i} $$ Multiply by the conjugate: $$ \frac{1}{-32 + 24i} \cdot \frac{-32 - 24i}{-32 - 24i} = \frac{-32 - 24i}{(-32)^2 + (24)^2} = \frac{-32 - 24i}{1024 + 576} = \frac{-32 - 24i}{1600} = \frac{-32}{1600} - \frac{24}{1600}i = -\frac{1}{50} - \frac{3}{200}i $$ The complex number is $-\frac{1}{50} - \frac{3}{200}i$ The conjugate of the complex number is $-\frac{1}{50} + \frac{3}{200}i$ Calculate the product with its conjugate: $$ (-\frac{1}{50} - \frac{3}{200}i)(-\frac{1}{50} + \frac{3}{200}i) = (-\frac{1}{50})^2 - (\frac{3}{200}i)^2 = \frac{1}{2500} + \frac{9}{40000} = \frac{16 + 9}{40000} = \frac{25}{40000} = \frac{1}{1600} $$

7. Solve A.1.g: $(-\frac{\sqrt{2}}{\frac{i}{2}})^{20}$ Simplify: $$ (-\frac{\sqrt{2}}{\frac{i}{2}})^{20} = (-\frac{2\sqrt{2}}{i})^{20} = (\frac{2\sqrt{2}}{-i})^{20} = (\frac{2\sqrt{2}}{-i} \cdot \frac{i}{i})^{20} = (\frac{2\sqrt{2}i}{-i^2})^{20} = (2\sqrt{2}i)^{20} = (2\sqrt{2})^{20} \cdot i^{20} $$ Note $i^{20} = (i^4)^5 = 1$, so we have: $$ (2\sqrt{2})^{20} = (2^{3/2})^{20} = 2^{30} = (2^{10})^3 = 1024^3 = 1073741824 $$ The complex number is 1073741824 The conjugate of the complex number is 1073741824 Calculate the product with its conjugate: $$ 1073741824 \cdot 1073741824 = 1152921504606846976 $$

8. Determine if the equalities are true or false A.2.1: $|(6+3i) + (4+2i)| = |6 + 3i| + |4 + 2i|$ $$ LHS: |(6+3i) + (4+2i)| = |10 + 5i| = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} $$ $$ RHS: |6 + 3i| + |4 + 2i| = \sqrt{6^2 + 3^2} + \sqrt{4^2 + 2^2} = \sqrt{36 + 9} + \sqrt{16 + 4} = \sqrt{45} + \sqrt{20} = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5} $$ The equality is TRUE.

9. Determine if the equalities are true or false A.2.2: $|3-2i| - |3+ 2i| = 4i$ $$ LHS: |3-2i| - |3+2i| = \sqrt{3^2 + (-2)^2} - \sqrt{3^2 + 2^2} = \sqrt{9 + 4} - \sqrt{9 + 4} = \sqrt{13} - \sqrt{13} = 0 $$ $$ RHS: 4i $$ Since $0 \neq 4i$, the equality is FALSE.

10. Determine if the equalities are true or false A.2.3: $|3 - 2i| < |3+ 2i|$ $$ LHS: |3 - 2i| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} $$ $$ RHS: |3 + 2i| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} $$ Since $\sqrt{13} \nless \sqrt{13}$, the equality is FALSE.

11. Determine if the equalities are true or false A.2.4: $|(3 – 2i) (3 + 2i)| = |3 - 2i||3 + 2i|$ $$ LHS: |(3 - 2i)(3 + 2i)| = |9 + 6i - 6i - 4i^2| = |9 + 4| = |13| = 13 $$ $$ RHS: |3 - 2i| |3 + 2i| = \sqrt{3^2 + (-2)^2} \sqrt{3^2 + 2^2} = \sqrt{13} \sqrt{13} = 13 $$ The equality is TRUE.

12. Demonstrate the conjugate property A.3.a: $\overline{z \cdot w} = \overline{z} \cdot \overline{w}$ Let $z = a + bi$ and $w = c + di$, where $a,b,c,d \in R$. Then $\overline{z} = a - bi$ and $\overline{w} = c - di$. $$ z \cdot w = (a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i $$ $$ \overline{z \cdot w} = \overline{(ac - bd) + (ad + bc)i} = (ac - bd) - (ad + bc)i $$ On the other hand: $$ \overline{z} \cdot \overline{w} = (a - bi)(c - di) = ac - adi - bci + bdi^2 = (ac - bd) - (ad + bc)i $$ Since $\overline{z \cdot w} = (ac - bd) - (ad + bc)i$ and $\overline{z} \cdot \overline{w} = (ac - bd) - (ad + bc)i$, the equality $\overline{z \cdot w} = \overline{z} \cdot \overline{w}$ is TRUE.

13. Demonstrate the conjugate property A.3.b: $\overline{z} = z$ if, and only if z is real. Let $z = a + bi$, where $a, b \in R$. The conjugate is $\overline{z} = a - bi$.

If $z$ is real, then $b = 0$, and $z = a$. Thus, $\overline{z} = a = z$. Conversely, if $\overline{z} = z$, then $a - bi = a + bi$, implies $-bi = bi$, and $2bi = 0$. Since $b$ is real, this implies $b = 0$. Thus $z = a$, which is real. Therefore, $\overline{z} = z$ if and only if $z$ is real.

14. Determine the values of x and y in R that make the following expression true A.4.a: $(5-2i) - 7 = x - (3 + yi)$ $$ (5 - 2i) - 7 = x - (3 + yi) \implies -2 - 2i = x - 3 - yi $$ $$ -2 - 2i = (x - 3) - yi $$ Equating real and imaginary parts: $$ -2 = x - 3 \implies x = 1 $$ $$ -2 = -y \implies y = 2 $$ Thus, $x = 1$ and $y = 2$.

15. Determine the values of x and y in R that make the following expression true A.4.b: $-6+ (3x + y)i = 3x + 7yi$ $$ -6+ (3x + y)i = 3x + 7yi $$ Equating real and imaginary parts: $$ -6 = 3x \implies x = -2 $$ $$ 3x + y = 7y \implies 3x = 6y \implies x = 2y $$ Since $x = -2$, $-2 = 2y \implies y = -1 $ Thus $x = -2$ and $y = -1$.

16. Solve the equation B.1.a: $\frac{1}{2x-1} = \frac{4}{8x-4}$ $$ \frac{1}{2x-1} = \frac{4}{8x-4} = \frac{4}{4(2x-1)} = \frac{1}{2x-1} $$ The original equation is $\frac{1}{2x-1} = \frac{1}{2x-1}$. So, the equation is satisfied for any $x$ except for when if $2x-1=0$. $ 2x-1 \neq 0$ so $x \neq \frac{1}{2} $. Answer is all real numbers except $\frac{1}{2}$.

17. Solve the equation B.1.b: $\sqrt{2x} - \frac{1}{\sqrt{2}} = \sqrt{8x}$ $$ \sqrt{2x} - \frac{1}{\sqrt{2}} = \sqrt{8x} = 2\sqrt{2x} \implies \sqrt{2x} - 2\sqrt{2x} = \frac{1}{\sqrt{2}} \implies -\sqrt{2x} = \frac{1}{\sqrt{2}} $$ $$ \sqrt{2x} = -\frac{1}{\sqrt{2}} $$ Since the square root cannot be negative, there is no solution.

18. Solve the equation B.1.c: $\frac{3}{2x+3} + \frac{5}{2x-3} = \frac{4x +6}{4x^{2}-9}$ Note that $4x^2-9=(2x+3)(2x-3)$ Therefore $$ \frac{3}{2x+3} + \frac{5}{2x-3} = \frac{3(2x-3) + 5(2x+3)}{(2x+3)(2x-3)} = \frac{6x - 9 + 10x + 15}{4x^2 - 9} = \frac{16x + 6}{4x^2 - 9} $$ The equation becomes: $$ \frac{16x + 6}{4x^2 - 9} = \frac{4x +6}{4x^2 - 9} $$ Therefore, $$ 16x + 6 = 4x + 6 $$ $$ 12x = 0$$ $$ x = 0 $$ Also, $2x+3 \neq 0$ and $2x - 3 \neq 0$, thus $x \neq \pm\frac{3}{2}$. So since $x=0$ is not $\pm\frac{3}{2}$, the solution is $x = 0$.

19. Solve the equation B.1.d: $\sqrt{10 + 3\sqrt{t}} = \sqrt{t}$ Square both sides $$ (\sqrt{10 + 3\sqrt{t}})^2 = (\sqrt{t})^2 $$ $$ 10 + 3\sqrt{t} = t$$ Let $u = \sqrt{t}$ then $u^2 = t$. The equation becomes: $$ 10 + 3u = u^2$$ $$ u^2 - 3u - 10 = 0$$ $$ (u-5)(u+2) = 0$$ So $u = 5$ or $u = -2$ Since $u = \sqrt{t}$, $u$ cannot be negative So $u=5$. So $5 = \sqrt{t}$, $t = 25$.

20. Solve the equation B.1.e: $3x^{\frac{2}{3}} + 4x^{\frac{1}{3}} - 4=0$ Let $y=x^{\frac{1}{3}}$, then $y^2 = x^{\frac{2}{3}}$ Thus, $3y^2 + 4y - 4 = 0$. $$ (3y - 2)(y + 2) = 0$$ So $y = \frac{2}{3}$ or $y = -2$ If $y = \frac{2}{3}$, $x^{\frac{1}{3}} = \frac{2}{3} \implies x = (\frac{2}{3})^3 = \frac{8}{27} $ If $y = -2$, $x^{\frac{1}{3}} = -2 \implies x = (-2)^3 = -8 $ Thus $x = \frac{8}{27}$ and $x = -8$