JEE Complex Numbers 2

Here's a trigonometric approach to the problem:

$z_{1}z_{2}=rr'(cos(\alpha)+isin(\alpha$ ) ) So $Im(z_{1}z_{2})=1 \Rightarrow rr'=csc(\alpha)$

And we know That: $|z_{1}|^2+|z_{2}|^2 \geq 2|z_{1}z_{2}|$

And: $|z_{1}z_{2}|=rr'$

So, we can write the following inequality:

$|z_{1}|^2+|z_{2}|^2+Re(z_{1}z_{2}) \geq 2|z_{1}z_{2}|+Re(z_{1}z_{2})$ $Re(z_{1}z_{2})=cot(\alpha)$ $2|z_{1}z_{2}|=2csc(\alpha)$

Now we are left with the expression: $2csc(\alpha)+cot(\alpha)$

All we do now is to take the derivative of the above expression with respect to $\alpha$ and find the minimum value of That expression wich is: $\boxed {\sqrt{3}}$

Write $z_1$ = $r_1$ $e^i$ $^A$ and $z_2$ = $r_2$ $e^i$ $^B$ using Eulers's form. So $Im(z_1z_2$ )= $r_1$ $r_2$ $sin(A+B)$ =1 From the second equation, we can find out the real part which will be $r_1$ $r_2$ $cos(A+B)$ ) So we have to find the minimum value of $(r_1)^2+(r_2)^2\pm\sqrt{(r_1r_2)^2-1}$ This comes out as $\sqrt{3}$ using partial derivatives

z1 = a +bi, z2 = c +di z1z2 = ac - bd+ (bc+ad)i

K = a^2 + b^2 + c^2 + d^2+ ac - bd, with bc +ad =1 K = (a^2 + c^2) + (b^2 + d^2) +ac - bd >= 2ac + 2bd + ac - bd (Cauchy formula) = 3ac + bd = 3a^2 + b^2 (c = a & d = b) >=2sqrt(3) a b = sqrt(3) (bc + ad =1 <=.>ab=1/2)