Question

could you tell me how would i solve this question
Q1 2-4i/4i

Q2 3-i/i

Q3 2y-3 + (4x+8)i =0

Answer 1

Answer on Question #52135, Math, Complex Analysis

Question: could you tell me how would i solve this question

Q1 2-4i/4i

Q2 3-i/i

Q3 $2y-3 + (4x+8)i = 0$

Answer:

Q1 $\frac{2 - 4i}{4i} = \frac{(2 - 4i) \times 4i}{4i \times 4i} = \frac{8i + 16}{-16} = -\frac{i + 2}{2} = -\frac{1}{2}i - 1$

Q2 $\frac{3 - i}{i} = \frac{(3 - i) \times i}{i \times i} = \frac{3i + 1}{-1} = -3i - 1$

Q3 $2y - 3 + (4x + 8)i = 0$

$$2y - 3 = 0 \quad 4x + 8 = 0$$ $$y = \frac{3}{2} \quad x = -2$$

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