The way we define the Logarithm is the inverse of the exponential.
If $z= re^{i\theta}$,then $ln(z)=ln(r)+i\theta$.
Now , a problem that is there is, with this equation alone , logarithm is multi-valued i.e for every z there are infinitely many values for the logarithm . For $\theta$ differing by a value of $2\pi$ , one will get the same value. It isn't injective. This leads us restricting its domain so that each point(z) corresponds to only one value. This is referred to as a branch cut. Check the image from Wiki,
So, for log(z), simply remove any ray joining 0 and infinity.(Restricting complete rotation(0 to $2\pi$!). Also, see in the graph what happens. The principle log removes one particular ray, the negative real axis, I guess. This is only a convention, I guess.
Now, coming back to your question, you want log(iz)(PS:-The $iz$ only changes the labels on the axes) to be analytic in some given region. All you need to do is to remove a ray(not in the given region) to make it analytic in the given region.
This is a standard problem in CA, but it is mispresented in the case of the OP.
The correct question (as it appears on Churchill and Brown for example) is:
If $\log(z)$ denotes the complex logarithm and $z\in\mathbb{C}$, show that it $\mathit{MAY}$ happen that $\log(z^2)\neq 2\log(z)$ (or with some other particular example which breaks the power rule...)
Not that it WILL happen, afortiori.
Like many commenters already said, this depends on the choice of branch of the complex function $\log$ one takes, but there's more to it than just that.
If the same branch is chosen for both cases, the result is equality. To dispel notions to the contrary, with Maple:
z0 := (1+I)^2;
z1 := 1+I;
CLog := proc (k, z) local theta, rho;
rho := op(1, polar(z)); theta := op(2, polar(z));
ln(rho)+I*(theta+2*k*Pi) end proc;
and using the same branch, we have:
CLog(0, z0);
ln(2)+(1/2*I)*Pi
2*CLog(0, z1);
ln(2)+(1/2*I)*Pi
Different pairs of $k\in\mathbb{Z}$ in the previous, will of course break equality.
There is, however, a different reason why equality may break: A different $\mathit{definition}$ of the complex function $\log$.
The standard definition used most often with the cut being the negative axis, is:
$$\log(k,z)=\ln(|z|)+(\arg(z)+2k\pi)i,\text{with:}\arg(z)\in(-\pi,\pi], k\in\mathbb{Z}$$
One, however, may define the complex $\log$ function to admit a branch cut starting from zero (where the function is not even defined) and extending out to infinity towards any direction one wants, so if one defines for example:
$$\log^*(k,z)=\ln(|z|)+(\arg(z)+2k\pi)i,\text{with:}\arg(z)\in(2\pi/p-2\pi,2\pi/p],k\in\mathbb{Z},p\in\mathbb{N}\setminus\{1\}$$
then this is a perfectly good definition for the complex $\log$ map, but will break many such equalities if one chooses $\arg$ lying on opposite sides of the cut depending on the complex values given, thus the inequality for specific examples.
Concluding, equality may break not only with different branches but also with different definitions of $\log$.
Best Answer
Since $$0.1x^2+x+3=0.1(x+5)^2+0.5>0$$ we have that $$|0.1x^2+2x+3|\ge 0.1x^2+2x+3>x>\log x$$ The equation has no real solution.