Question

For the following question, enter the correct numerical value up to TWO decimal places. (If the numerical value has more than two decimal places, round-off the value to TWO decimal places.) If \(\alpha,\beta,\gamma\) and \(a,b,c\) are complex numbers such that \[ \frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}=1+i \quad \text{and} \quad \frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}=0, \] then the value of \[ \frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}=____.\, i \]

Hint:

Useful identities:
\((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\)
If \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\), then \(xy+yz+zx=0\) These greatly simplify complex number problems.

Answer 1

Correct Answer: 2

Step 1: Let \[ x=\frac{\alpha}{a},\quad y=\frac{\beta}{b},\quad z=\frac{\gamma}{c}. \] Then the given conditions become: \[ x+y+z=1+i \qquad (1) \] \[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \qquad (2) \]
Step 2: From equation (2): \[ \frac{xy+yz+zx}{xyz}=0 \Rightarrow xy+yz+zx=0 \qquad (3) \]
Step 3: Use the identity: \[ (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) \]
Step 4: Substitute from (1) and (3): \[ (1+i)^2=x^2+y^2+z^2+2(0) \] \[ x^2+y^2+z^2=(1+i)^2 \]
Step 5: Evaluate: \[ (1+i)^2=1+2i+i^2=2i \]
Step 6: Hence, \[ \frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}=2i \] Writing the numerical value up to two decimal places: \[ \boxed{2.00\,i} \]