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PROBLEM

ii. Solve 12+2i\frac{1}{\sqrt{2}+\sqrt{2} i} in the form a+bia+b i.
Selesaikan 12+2i\frac{1}{\sqrt{2}+\sqrt{2} i} dalam bentuk a+bia+b i.

STEP 1

1. We are given a complex number in the form of a fraction.
2. The goal is to express this complex number in the standard form a+bia + bi.
3. To achieve this, we will multiply the numerator and the denominator by the conjugate of the denominator.

STEP 2

1. Identify and multiply by the conjugate of the denominator.
2. Simplify the expression to obtain the form a+bia + bi.

STEP 3

Identify the conjugate of the denominator 2+2i\sqrt{2} + \sqrt{2}i. The conjugate is 22i\sqrt{2} - \sqrt{2}i.
Multiply both the numerator and the denominator by this conjugate:
12+2i×22i22i\frac{1}{\sqrt{2}+\sqrt{2} i} \times \frac{\sqrt{2} - \sqrt{2} i}{\sqrt{2} - \sqrt{2} i}

STEP 4

Simplify the denominator using the difference of squares formula:
(2+2i)(22i)=(2)2(2i)2(\sqrt{2} + \sqrt{2}i)(\sqrt{2} - \sqrt{2}i) = (\sqrt{2})^2 - (\sqrt{2}i)^2 Calculate each term:
(2)2=2and(2i)2=2i2=2(\sqrt{2})^2 = 2 \quad \text{and} \quad (\sqrt{2}i)^2 = 2i^2 = -2 Thus, the denominator becomes:
2(2)=2+2=42 - (-2) = 2 + 2 = 4

STEP 5

Simplify the numerator:
1×(22i)=22i1 \times (\sqrt{2} - \sqrt{2}i) = \sqrt{2} - \sqrt{2}i

SOLUTION

Combine the results to express the fraction in the form a+bia + bi:
22i4=2424i\frac{\sqrt{2} - \sqrt{2}i}{4} = \frac{\sqrt{2}}{4} - \frac{\sqrt{2}}{4}i Thus, the expression in the form a+bia + bi is:
a=24,b=24a = \frac{\sqrt{2}}{4}, \quad b = -\frac{\sqrt{2}}{4}

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