Question9. $\frac{2+i}{3-2 i}$
10. $\frac{2+i-2 i^{3}}{2 i^{3}+4}$
11. $\frac{2+2 \sqrt{3}}{1-\sqrt{3}}$

Studdy Solution

STEP 1

1. The expressions involve complex numbers, where $i$ is the imaginary unit with the property $i^2 = -1$ .
2. The expressions can be simplified using algebraic manipulations, including the rationalization of denominators.
3. The final expressions should be in the form $a+bi$ , where $a$ and $b$ are real numbers.

STEP 2

1. Simplify the expression $\frac{2+i}{3-2i}$ by rationalizing the denominator.
2. Simplify the expression $\frac{2+i-2i^{3}}{2i^{3}+4}$ by simplifying the numerator and denominator.
3. Simplify the expression $\frac{2+2\sqrt{3}}{1-\sqrt{3}}$ by rationalizing the denominator.

STEP 3

To simplify $\frac{2+i}{3-2i}$ , multiply the numerator and the denominator by the complex conjugate of the denominator, $3+2i$ .
$\frac{2+i}{3-2i} \cdot \frac{3+2i}{3+2i}$

STEP 4

Expand the numerator and the denominator:
$\frac{(2+i)(3+2i)}{(3-2i)(3+2i)}$

STEP 5

Calculate the products in the numerator and the denominator:
$\text{Numerator: } (2+i)(3+2i) = 6 + 4i + 3i + 2i^2 = 6 + 7i - 2 = 4 + 7i$ $\text{Denominator: } (3-2i)(3+2i) = 9 - 4i^2 = 9 + 4 = 13$

STEP 6

Write the simplified expression:
$\frac{4 + 7i}{13} = \frac{4}{13} + \frac{7i}{13}$ Therefore, the simplified form of $\frac{2+i}{3-2i}$ is:
$\frac{4}{13} + \frac{7i}{13}$

STEP 7

Simplify $\frac{2+i-2i^3}{2i^3+4}$ by recognizing that $i^3 = -i$ , and substitute $i^3$ with $-i$ .
$\frac{2+i-2(-i)}{2(-i)+4} = \frac{2+i+2i}{-2i+4}$

STEP 8

Combine like terms in the numerator and denominator:
$\frac{2+3i}{4-2i}$

STEP 9

Rationalize the denominator by multiplying the numerator and the denominator by the complex conjugate of the denominator, $4+2i$ .
$\frac{2+3i}{4-2i} \cdot \frac{4+2i}{4+2i}$

STEP 10

Expand the products in the numerator and the denominator:
$\frac{(2+3i)(4+2i)}{(4-2i)(4+2i)}$

STEP 11

Calculate the products in the numerator and the denominator:
$\text{Numerator: } (2+3i)(4+2i) = 8 + 4i + 12i + 6i^2 = 8 + 16i - 6 = 2 + 16i$ $\text{Denominator: } (4-2i)(4+2i) = 16 - 4i^2 = 16 + 4 = 20$

STEP 12

Write the simplified expression:
$\frac{2 + 16i}{20} = \frac{2}{20} + \frac{16i}{20} = \frac{1}{10} + \frac{4i}{5}$ Therefore, the simplified form of $\frac{2+i-2i^3}{2i^3+4}$ is:
$\frac{1}{10} + \frac{4i}{5}$

STEP 13

Simplify $\frac{2+2\sqrt{3}}{1-\sqrt{3}}$ by rationalizing the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, $1+\sqrt{3}$ .
$\frac{2+2\sqrt{3}}{1-\sqrt{3}} \cdot \frac{1+\sqrt{3}}{1+\sqrt{3}}$

STEP 14

Expand the products in the numerator and the denominator:
$\frac{(2+2\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}$

STEP 15

Calculate the products in the numerator and the denominator:
$\text{Numerator: } (2+2\sqrt{3})(1+\sqrt{3}) = 2 + 2\sqrt{3} + 2\sqrt{3} + 6 = 8 + 4\sqrt{3}$ $\text{Denominator: } (1-\sqrt{3})(1+\sqrt{3}) = 1 - (\sqrt{3})^2 = 1 - 3 = -2$

STEP 16

Write the simplified expression:
$\frac{8 + 4\sqrt{3}}{-2} = \frac{8}{-2} + \frac{4\sqrt{3}}{-2} = -4 - 2\sqrt{3}$ Therefore, the simplified form of $\frac{2+2\sqrt{3}}{1-\sqrt{3}}$ is:
$-4 - 2\sqrt{3}$
Solution:
1. $\frac{2+i}{3-2i}$ simplifies to $\frac{4}{13} + \frac{7i}{13}$ .
2. $\frac{2+i-2i^3}{2i^3+4}$ simplifies to $\frac{1}{10} + \frac{4i}{5}$ .
3. $\frac{2+2\sqrt{3}}{1-\sqrt{3}}$ simplifies to $-4 - 2\sqrt{3}$ .

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