I give up. I can't solve the problem given, I think more information is needed beyond what is in the problem statement, and I wouldn't be saying that if I had not hacked away at it and wound up at this point. To begin with, the problem is as follows.
We have voltage generator \$E=2\sqrt{7} \mbox{ } V\$ with angular frequency \$\omega=10^6 \mbox{ } s^{-1}\$ and internal resistance \$R_g=0.5\sqrt{3} \mbox{ } k\Omega\$ connected to parallel connection of impedance \$Z\$ and coil \$L\$. Current is \$I=I_1=I_2=4 \mbox{ } mA\$. Calculate complex value of \$\underline{Z}\$ and inductivity of \$L\$.
My claim is that this is unsolvable. I owe a little explanation for for my claim before I change the problem and solve something different. Basically, the fact that \$\underline{Z}\$ and \$L\$ are unknown gives 3 unknowns. Combined with the power factor of the circuit, this gives 4 real unknowns. You can do mesh analysis or node analysis and find that you will have 2 complex equations, minus one reference. You're one short.
Here is what I would add:
Assume that the magnitude of \$I_1\$ and \$I_2\$ are equal.
The only way I know to do this is to use the answer given in the problem, so now that I have that out of the way I'll hack away at this. I'll introduce only \$Z_{e}\$, which is the combined impedance of the 2 parallel components. I might also forget some of the vector bars, forgive me please. Start at the voltage source and note the following, using the general \$|V|=|I| |Z|\$ property.
$$|E| = |I| |Z_g+Z_e|$$
$$|Z_g+Z_e| = \frac{ |E| }{|I|} = 500 \sqrt{7}$$
Now I'll define my reference and follow through the voltage a bit. The notation I use is \$U_1\$ for that obvious voltage point after the resistor. I'm using \$-\psi\$ for the current angle because I already know it's a net inductive circuit, which is just from knowledge of the solution.
$$ E = 2 \sqrt{7} \angle 0 $$
$$ I = \frac{1}{250} \angle -\psi$$
$$ U_1 = E - R I = 2 \sqrt{7} - 2 \sqrt{3} \angle -\psi$$
I need to write the equation for the equivalent inductance.
$$ Z_e = \frac{1}{ \frac{1}{Z} + \frac{1}{j \omega L} } $$
Anyway, I'll just skip some steps and write the values. I hope to come back and put more in later. Sorry about the lack of actual circuit analysis in this answer.
$$ \psi = arctan( \frac{1}{3 \sqrt{3} } )$$
$$ Z = 250 \angle -\frac{\pi}{3} $$
$$ Z_e = 250 \angle \frac{\pi}{3} $$
$$ I_1 = \frac{1}{250} \angle arctan( \frac{2}{\sqrt{3}} )$$
$$ I_1 = \frac{1}{250} \angle -arctan( \frac{5 \sqrt{7}}{\sqrt{21}} )$$
It's already redundant to say this, but these numbers give the \$Z=250(\sqrt{3}-j)\$ and \$L=0.5 mH\$. It would also work to say that Z is a resistor of \$250 \sqrt{3} \Omega \$ in series with a \$ 4 nF\$ capacitor.
I think this was a bad question, and I hope I've given enough breadcrumbs of a consistent answer for your to prove this to someone else. Maybe I'm wrong, but if my current analysis is right, I would hate to have for anyone to be given this on a test.
Your total impedance is correct. As you can see cos(-19.653)=0.94 which is below 0.95.
If you calculate the total impedance when the 10 Ohms resistor is shorted you'll find 10<0, i.e. PF=1.
Therefore you can increase the PF by placing a small impedance accross the 10 Ohms resistor, and then the equivalent impedance modulus of both components will be below 10 Ohms (and below the small impedance). To get PF=1 you would place a really small reactance, whether it is an inductor or a capacitor.
To get PF=0.95 you can also succeed either with an inductor or with a capacitor.
Let's solve this using a capacitor. One way to do it is to calculate the input impedance (seen from the source).
If I is the current flowing from the source, the source voltage can be written as
$$ V=-j10·I + \frac { I·(j20+Zx) } {20 +j20+Zx }·20$$
The right term is the voltage in resistor 20 obtained by multiplying the current in this resistor (current divisor formula) by 20. And Zx is parallel between 10 Ohms and unknown reactance \$Zx=\frac {-jXc·10} {10-jXc}\$.
Therefore the Z, input impedance is
$$ Z=-j10 + \frac { (j20+\frac {-jXc·10} {10-jXc}) } {20 +j20+\frac {-jXc·10} {10-jXc} }·20$$
Using algebra (or some symbolic math package) you can get real part and imaginary part of Z.
Dividing Imaginary part and real part, and simplifying, you get:
$$\frac {\Im (Z)}{\Re{(Z)}}=-2.5·\frac {Xc^2} {7Xc^2-40Xc+400}$$
This division must be equal to $$-\tan(\arccos(0.95)\approx -0.329$$
Solving for Xc, you get 2 solutions, taking the positive one, \$Xc=8.821 \Omega \$. The capacitance is \$C \approx 361 \mu F\$
Best Answer
Both these equations are correct because \$j = \sqrt{-1} \$
We have
$$Z_{cap} = \dfrac{1}{j \ \omega \ C} = \dfrac{1}{\sqrt{-1} \ \omega \ C} = \dfrac{1}{\sqrt{-1} \ \omega \ C} \cdot \dfrac{\sqrt{-1}}{\sqrt{-1}} = - \dfrac{\sqrt{-1}}{\omega \ C} = - \dfrac{j}{\omega \ C}$$