Given that `(2 sqrt3 cos 30^(@) - 2i sin 30^(@))/(sqrt2 (cos 45^(@) + isin 45^(@)))=A + Bi`, find the values of A and B.

Answer

To solve the given problem, we will follow these steps:

**Step 1: Substitute the trigonometric values.**

We know the values of the trigonometric functions:
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 30^\circ = \frac{1}{2} \)
- \( \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \)

Substituting these values into the expression:

\[ \frac{2\sqrt{3} \cos 30^\circ - 2i \sin 30^\circ}{\sqrt{2} (\cos 45^\circ + i \sin 45^\circ)} = \frac{2\sqrt{3} \cdot \frac{\sqrt{3}}{2} - 2i \cdot \frac{1}{2}}{\sqrt{2} \left(\frac{1}{\sqrt{2}} + i \cdot \frac{1}{\sqrt{2}}\right)} \]

**Step 2: Simplify the numerator.**

Calculating the numerator:

\[ 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 3 \]
\[ -2i \cdot \frac{1}{2} = -i \]

Thus, the numerator simplifies to:

\[ 3 - i \]

**Step 3: Simplify the denominator.**

Calculating the denominator:

\[ \sqrt{2} \left(\frac{1}{\sqrt{2}} + i \cdot \frac{1}{\sqrt{2}}\right) = \sqrt{2} \cdot \frac{1}{\sqrt{2}} + i \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 + i \]

**Step 4: Combine the results.**

Now we have:

\[ \frac{3 - i}{1 + i} \]

**Step 5: Multiply by the conjugate of the denominator.**

To simplify this expression, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{(3 - i)(1 - i)}{(1 + i)(1 - i)} = \frac{(3 - i)(1 - i)}{1^2 - i^2} = \frac{(3 - i)(1 - i)}{1 - (-1)} = \frac{(3 - i)(1 - i)}{2} \]

**Step 6: Expand the numerator.**

Expanding the numerator:

\[ (3 - i)(1 - i) = 3 \cdot 1 - 3i - i + i^2 = 3 - 3i - i - 1 = 2 - 4i \]

So we have:

\[ \frac{2 - 4i}{2} = 1 - 2i \]

**Step 7: Identify A and B.**

From the expression \(1 - 2i\), we can identify:

\[ A = 1, \quad B = -2 \]

Thus, the values of \(A\) and \(B\) are:

\[ \boxed{1} \quad \text{and} \quad \boxed{-2} \]

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