Question

micah richard
advanced math
homework 5

1. \\(\sqrt{-121}\\)

\\(\sqrt{121}\cdot i = 11i\\)

1. \\(\sqrt{\frac{-81}{4}}\\)

\\(\sqrt{\frac{81}{4}}\cdot i\\)

1. \\(i^{59}\\)
2. \\((-3i^{16})^3\\)
3. \\((i\sqrt4{6})^2\cdot (-4i)^3\\)
4. \\((8 + 5i) - (6 + 4i)\\)
5. \\((-16 - 3i) + (-3 - 8i)\\)
6. \\(-i(8 + 2i) - 4i(10 - 3i\\)

Explanation:

Step1: Rewrite sqrt of negative number

$\sqrt{-121} = \sqrt{121} \cdot i$

Step2: Simplify the radical

$\sqrt{121} \cdot i = 11i$

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Step1: Rewrite sqrt of negative fraction

$\sqrt{-\frac{81}{4}} = \sqrt{\frac{81}{4}} \cdot i$

Step2: Simplify the radical

$\sqrt{\frac{81}{4}} \cdot i = \frac{9}{2}i$

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

$59 = 4 \cdot 14 + 3$, so $i^{59}=i^{4\cdot14 + 3}=(i^4)^{14} \cdot i^3$

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

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

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Step1: Simplify $i^{16}$ using $i^4=1$

$i^{16}=(i^4)^4=1^4=1$

Step2: Compute the power

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

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Step1: Simplify each complex term

$(i\sqrt{6})^2 = i^2 \cdot (\sqrt{6})^2 = -1 \cdot 6 = -6$
$(-4i)^3 = (-4)^3 \cdot i^3 = -64 \cdot (-i) = 64i$

Step2: Multiply the results

$-6 \cdot 64i = -384i$

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Step1: Distribute the negative sign

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

Step2: Combine like terms

$(8-6)+(5i-4i) = 2+i$

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Step1: Remove parentheses and combine terms

$(-16-3i)+(-3-8i) = -16-3-3i-8i$

Step2: Simplify like terms

$-19-11i$

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Step1: Distribute the coefficients

$-i(8+2i)-4i(10-3i) = -8i-2i^2-40i+12i^2$

Step2: Substitute $i^2=-1$ and combine terms

$-8i-2(-1)-40i+12(-1) = -8i+2-40i-12$

Step3: Combine like terms

$(-8i-40i)+(2-12) = -48i-10 = -10-48i$

Answer:

1. $11i$
2. $\frac{9}{2}i$
3. $-i$
4. $-27$
5. $-384i$
6. $2+i$
7. $-19-11i$
8. $-10-48i$