Express the result in the form x+iy, where x,y are real number `i=sqrt(-1)`:
(i) `(2-3i)/(4-i)`
(ii) `(2+3i)/(-5-4i)`
(iii) `(1+i)/(3+i)`
(iv) `(3+2i)/(4-3i)`

Answer

To express the given complex numbers in the form \( x + iy \), where \( x \) and \( y \) are real numbers, we will follow a systematic approach for each part.

### (i) \(\frac{2-3i}{4-i}\)

**Step 1:** Multiply the numerator and denominator by the conjugate of the denominator.

\[ \frac{2-3i}{4-i} \cdot \frac{4+i}{4+i} = \frac{(2-3i)(4+i)}{(4-i)(4+i)} \]

**Step 2:** Calculate the denominator.

\[ (4-i)(4+i) = 4^2 - (-1) = 16 + 1 = 17 \]

**Step 3:** Calculate the numerator.

\[ (2-3i)(4+i) = 2 \cdot 4 + 2 \cdot i - 3i \cdot 4 - 3i \cdot i = 8 + 2i - 12i - 3(-1) = 8 - 10i + 3 = 11 - 10i \]

**Step 4:** Combine the results.

\[ \frac{11 - 10i}{17} = \frac{11}{17} - \frac{10}{17}i \]

Thus, in the form \( x + iy \):

\[ x = \frac{11}{17}, \quad y = -\frac{10}{17} \]

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### (ii) \(\frac{2+3i}{-5-4i}\)

**Step 1:** Factor out the negative from the denominator.

\[ \frac{2+3i}{-5-4i} = -\frac{2+3i}{5+4i} \]

**Step 2:** Multiply by the conjugate of the denominator.

\[ -\frac{2+3i}{5+4i} \cdot \frac{5-4i}{5-4i} = -\frac{(2+3i)(5-4i)}{(5+4i)(5-4i)} \]

**Step 3:** Calculate the denominator.

\[ (5+4i)(5-4i) = 5^2 - (4i)^2 = 25 + 16 = 41 \]

**Step 4:** Calculate the numerator.

\[ (2+3i)(5-4i) = 10 - 8i + 15i - 12(-1) = 10 + 7i + 12 = 22 + 7i \]

**Step 5:** Combine the results.

\[ -\frac{22 + 7i}{41} = -\frac{22}{41} - \frac{7}{41}i \]

Thus, in the form \( x + iy \):

\[ x = -\frac{22}{41}, \quad y = -\frac{7}{41} \]

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### (iii) \(\frac{1+i}{3+i}\)

**Step 1:** Multiply by the conjugate of the denominator.

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

**Step 2:** Calculate the denominator.

\[ (3+i)(3-i) = 3^2 - (i)^2 = 9 + 1 = 10 \]

**Step 3:** Calculate the numerator.

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

**Step 4:** Combine the results.

\[ \frac{4 + 2i}{10} = \frac{4}{10} + \frac{2}{10}i = \frac{2}{5} + \frac{1}{5}i \]

Thus, in the form \( x + iy \):

\[ x = \frac{2}{5}, \quad y = \frac{1}{5} \]

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### (iv) \(\frac{3+2i}{4-3i}\)

**Step 1:** Multiply by the conjugate of the denominator.

\[ \frac{3+2i}{4-3i} \cdot \frac{4+3i}{4+3i} = \frac{(3+2i)(4+3i)}{(4-3i)(4+3i)} \]

**Step 2:** Calculate the denominator.

\[ (4-3i)(4+3i) = 4^2 - (3i)^2 = 16 + 9 = 25 \]

**Step 3:** Calculate the numerator.

\[ (3+2i)(4+3i) = 12 + 9i + 8i + 6(-1) = 12 + 17i - 6 = 6 + 17i \]

**Step 4:** Combine the results.

\[ \frac{6 + 17i}{25} = \frac{6}{25} + \frac{17}{25}i \]

Thus, in the form \( x + iy \):

\[ x = \frac{6}{25}, \quad y = \frac{17}{25} \]

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