Question

If a,b,c, $a_1,b_1,c_1$ are non zero complex numbers satisfying $\frac{a}{a_1}+\frac{b}{b_1}+\frac{c}{c_1}=1+i$ and $\frac{a_1}{a}+\frac{b_1}{b}+\frac{c_1}{c}=0$, then $\frac{a^2}{a_1^2}+\frac{b^2}{b_1^2}+\frac{c^2}{c_1^2}$ is equal to

• A. 2+2i
• B. 2i
• C. 2-2i
• D. 2

Answer 1

Answer: (B)

2i

Answer 2

Let

$x=\frac{a}{a_1}, y=\frac{b}{b_1}, z=\frac{c}{c_1}$.

Then, the given equations become:

$x+y+z=1+i$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$.

The second equation can be rewritten as:

$\frac{xy+yz+zx}{xyz}=0 \implies xy+yz+zx=0$.

We need to find:

$x^2+y^2+z^2$.

Recall the identity:

$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$.

Since $xy+yz+zx=0$, we have:

$x^2+y^2+z^2=(x+y+z)^2=(1+i)^2$.

Now, compute $(1+i)^2$:

$(1+i)^2=1+2i+i^2=1+2i-1=2i$.

Thus, $\frac{a^2}{a_1^2}+\frac{b^2}{b_1^2}+\frac{c^2}{c_1^2}=2i$.