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micah richard advanced math homework 5 1. \\(\\sqrt{-121}\\) 17. \\((-1…

Question

micah richard
advanced math
homework 5

  1. \\(\sqrt{-121}\\) 17. \\((-1 - 4i)(11 + 6i)\\)
  2. \\(\sqrt{\frac{-81}{4}}\\)
  3. \\(i^{59}\\) 19. \\((2 - 3i)^2\\)
  4. \\((-3i^{16})^3\\) 21. \\(\frac{-5 - 2i}{6i}\\)
  5. \\((i\sqrt4{6})^2 \cdot (-4i)^3\\) 23. \\(\frac{7i}{6 + 2i}\\)
  6. \\((8 + 5i) - (6 + 4i)\\)

Explanation:

Step1: Rewrite using imaginary unit $i$

$\sqrt{-121} = \sqrt{121 \times (-1)} = \sqrt{121} \times \sqrt{-1}$

Step2: Simplify radicals and substitute $i=\sqrt{-1}$

$= 11i$

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Step1: Rewrite using imaginary unit $i$

$\sqrt{\frac{-81}{4}} = \sqrt{\frac{81}{4} \times (-1)} = \sqrt{\frac{81}{4}} \times \sqrt{-1}$

Step2: Simplify radicals and substitute $i=\sqrt{-1}$

$= \frac{9}{2}i$

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Step1: Use cyclicity of $i$ ($i^4=1$)

$i^{59} = i^{4\times14 + 3} = (i^4)^{14} \times i^3$

Step2: Substitute $i^4=1, i^3=-i$

$= 1^{14} \times (-i) = -i$

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Step1: Expand the power of complex term

$(-3i^{10})^3 = (-3)^3 \times (i^{10})^3$

Step2: Simplify $i^{10}$ using $i^4=1$

$i^{10}=i^{4\times2+2}=(i^4)^2\times i^2=1^2\times(-1)=-1$

Step3: Compute final value

$= -27 \times (-1)^3 = -27 \times (-1) = 27$

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Step1: Simplify each complex factor

$(i^4\sqrt{6})^2 = (i^4)^2 \times (\sqrt{6})^2$, $(-4i)^3=(-4)^3\times i^3$

Step2: Substitute $i^4=1, i^3=-i$

$(1)^2 \times 6 = 6$; $-64 \times (-i)=64i$

Step3: Multiply the two results

$6 \times 64i = 384i$

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Step1: Distribute the subtraction

$(8+5i)-(6+4i) = 8-6 + 5i-4i$

Step2: Combine like terms

$= 2 + i$

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Step1: Use FOIL method to multiply

$(-1-4i)(11+6i) = (-1)(11) + (-1)(6i) + (-4i)(11) + (-4i)(6i)$

Step2: Compute each product

$= -11 -6i -44i -24i^2$

Step3: Substitute $i^2=-1$ and combine like terms

$= -11 -50i -24(-1) = -11+24-50i = 13-50i$

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Step1: Expand using square of binomial

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

Step2: Compute each term

$= 4 -12i +9i^2$

Step3: Substitute $i^2=-1$ and simplify

$= 4-12i+9(-1) = 4-9-12i = -5-12i$

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Step1: Multiply numerator/denominator by $-i$

$\frac{-5-2i}{6i} \times \frac{-i}{-i} = \frac{(-5-2i)(-i)}{6i(-i)}$

Step2: Expand numerator and denominator

Numerator: $5i + 2i^2$; Denominator: $-6i^2$

Step3: Substitute $i^2=-1$ and simplify

$= \frac{5i + 2(-1)}{-6(-1)} = \frac{-2+5i}{6} = -\frac{1}{3} + \frac{5}{6}i$

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Step1: Multiply numerator/denominator by conjugate $6-2i$

$\frac{7i}{6+2i} \times \frac{6-2i}{6-2i} = \frac{7i(6-2i)}{6^2-(2i)^2}$

Step2: Expand numerator and denominator

Numerator: $42i -14i^2$; Denominator: $36-4i^2$

Step3: Substitute $i^2=-1$ and simplify

$= \frac{42i -14(-1)}{36-4(-1)} = \frac{14+42i}{40} = \frac{7}{20} + \frac{21}{20}i$

Answer:

  1. $11i$
  2. $\frac{9}{2}i$
  3. $-i$
  4. $27$
  5. $384i$
  6. $2+i$
  7. $13-50i$
  8. $-5-12i$
  9. $-\frac{1}{3} + \frac{5}{6}i$
  10. $\frac{7}{20} + \frac{21}{20}i$