Current divider usage
Here's your schematic:



simulate this circuit &ndash; Schematic created using CircuitLab
(By the way, if you plan to use this site again you should learn to use the included schematic editor. It automatically numbers the parts for you and that provides us with an easier, more specific way to address ourselves to your schematic and questions. It's not perfect. As you can see above, it insists on adding &quot;F&quot; and &quot;H&quot; where it's not wanted. But we can all live with these warts.)
For more explanation about why I laid this out as I did, see Note below.
I completely agree with your approach in finding the total \$Z\$ and the total \$I\$ for the circuit. You have \$R_1\$ in series with \$L_1\$ in series with the parallel impedance of \$R_2\$ and \$C_1+L_2\$. To compute \$I_{_\text{TOTAL}}=\frac{V_{_\text{S}}}{Z_{_\text{TOTAL}}}\$, you need to find \$Z_{_\text{TOTAL}}\$:
$$\begin{align*}
Z_{_\text{TOTAL}}&amp;=Z_1+\left(R_2 \mid\mid Z_2\right)\\&amp;=\left(4\:\Omega+j\,20\:\Omega\right) + \left(16\:\Omega \mid\mid \left[-j\,14\:\Omega+25\,j\:\Omega\right]\right)\\
&amp;=9.13527851 + j\,27.469496\Leftrightarrow 28.9486878 \:\angle\: 71.6048952^\circ\\&amp;\therefore\\I_{_\text{TOTAL}} &amp;=0.130811304 - j\,0.393345488\Leftrightarrow 0.414526561 \:\angle\: -71.6048952^\circ
\end{align*}$$
You found similar results. So that's good.
I think where you went wrong (and this is offered with respect, because I think you were actually pretty smart here -- just that you got it backwards) is that you constructed two ratios, \$\frac{Z_2}{R_2\mid\mid Z_2}\$ and \$\frac{R_2}{R_2\mid\mid Z_2}\$, as a means of dividing up the total current in the circuit.
But what I think you forgot here (the thing you got backwards, so to speak) is that you really wanted the ratio of each branch admittance to the total admittance. This is why I think your brain is working really well. It's just that you forgot to realize to first convert the impedences to admittances. So you were very close.
Let's see what happens when we apply your concept but instead use the correct objects, namely the admittances:
$$\begin{align*}
i_1&amp;=I_{_\text{TOTAL}}\cdot \frac{\frac1{R_2}}{\frac1{R_2\mid\mid Z_2}}=I_{_\text{TOTAL}}\cdot \frac{R_2\mid\mid Z_2}{R_2}\\\\&amp;=0.225615314 - j\,0.0651777575\Leftrightarrow 0.234841245 \:\angle\: -16.1134182^\circ\\\\
i_2&amp;=I_{_\text{TOTAL}}\cdot \frac{\frac1{Z_2}}{\frac1{R_2\mid\mid Z_2}}=I_{_\text{TOTAL}}\cdot \frac{R_2\mid\mid Z_2}{Z_2}\\\\&amp;=-0.0948040109 - j\,0.32816773\Leftrightarrow 0.341587265 \:\angle\: -106.113418^\circ
\end{align*}$$
And those are the right answers.
Your basic idea was on the right track. You just got mixed up about the objects upon which your idea applies.
We've all made such mistakes and I think you are going to do just fine on this stuff. And this kind of question you wrote is perhaps the classic kind of student question we should see presented here. You had a clear description of the problem you were working on and you showed all of your work. In addition, you showed enough about how you applied your thinking so that we could work out where you went wrong. If every student question were like this, I believe this site would find far more joy and readiness in helping answer them. So +1 for that!
Note
I've arranged things so that the divider in a way that eliminates having to add the source voltage. It's not necessary to see it or to see the wires busing power between it and the parts; and those wires just get in the way of seeing the remaining important bits. Also, I've arranged things so that we can use a conventional schematic drafting sheet of paper where currents flow from top to bottom (yes, I know this is AC) and where signal flows from left to right. (In this case, the schematic doesn't have a signal to worry about.) If you consistently draft your schematics to follow rules like this, you'll find them to be more understandable.
I like to think of it like a curtain or sheet where currents flowing from top to bottom are like the flow of a river downstream. Signals then propagate across that river from one side to the other side. These signals require that flow in order to operate, so to speak. It's a picturesque way of seeing things and it pretty much always works better than other alternatives. But there are times, such as with some power supply schematics, where the busing around of power is the important thing to show and highlight to make the schematic better understood. So it's not a hard and fast law. The main point is to have the schematic communicate with others and to do that you need to emphasize the parts and sections to which attention must be drawn and to avoid wasting &quot;ink&quot; on stuff that doesn't matter.