You are correct; set $x=a+i b$, $y=c+id$, replace in the equations and isolate the real and imaginary parts (setting them equal to zero).
Doing so, you get $$a^2-14 a-b^2+c^2-4 c-d^2-47=0 \tag 1$$ $$2 a b-14 b+2 c d-4 d=0\tag 2$$ $$a^2-22 a-b^2+c^2-10 c-d^2+71=0\tag 3$$ $$2 a b-22 b+2 c d-10 d=0\tag 4$$ You could solve these four equations using Newton-Raphson method.
What I found amazing is that the problem has analytical solutions.
$(1)-(3)$ leads to $$8 a+6 c-118=0 \tag 5$$ Similarly $(2)-(4)$ leads to $$8 b+6 d=0\tag 6$$ So, using $c=\frac{59-4 a}{3} $ and $d=-\frac{4 b}{3}$ and replacing we are let with $$a^2-22 a-b^2+94=0\tag 7$$ $$a b-11 b=0\tag 8$$ The last equation shows that either $a=11$ or $b=0$.
If $a=11$, $(7)$ gives $b^2+27=0$ that is to say $b=\pm 3 i \sqrt{3}$
If $b=0$, $(7)$ gives $a^2-22 a+94$ that is to say $a=11 \pm 3 \sqrt{3}$ For each pair, compute the corresponding $c$ and $d$ and the four solutions are $$\left\{a= 11,b= -3 i \sqrt{3},c= 5,d= 4 i \sqrt{3}\right\}$$ $$\left\{a=
11,b= 3 i \sqrt{3},c= 5,d= -4 i \sqrt{3}\right\}$$ $$\left\{a= 11+3
\sqrt{3},b= 0,c= 5-4 \sqrt{3},d= 0\right\}$$ $$\left\{a= 11-3 \sqrt{3},b=
0,c= 5+4 \sqrt{3},d= 0\right\}$$
Well, we have:
$$\exp\left(\frac{x^2}{4\cdot\text{v}\cdot t}\right)=1+\frac{x^2}{2\cdot\text{v}\cdot t}\tag1$$
Now, we know that we can write:
$$\exp\left(\alpha\right)=\sum_{\text{n}=0}^\infty\frac{\alpha^\text{n}}{\text{n}!}=\frac{\alpha^0}{0!}+\frac{\alpha^1}{1!}+\frac{\alpha^2}{2!}+\dots=$$
$$1+\alpha+\frac{\alpha^2}{2}+\dots\tag2$$
So, for equation $(1)$ we can write:
$$1+\frac{x^2}{4\cdot\text{v}\cdot t}+\frac{1}{2}\cdot\left(\frac{x^2}{4\cdot\text{v}\cdot t}\right)^2+\dots=1+\frac{x^2}{2\cdot\text{v}\cdot t}\tag3$$
Using the aproximation of three terms we have:
$$1+\frac{x^2}{4\cdot\text{v}\cdot t}+\frac{1}{2}\cdot\left(\frac{x^2}{4\cdot\text{v}\cdot t}\right)^2\approx1+\frac{x^2}{2\cdot\text{v}\cdot t}\space\Longleftrightarrow\space$$
$$x\approx0\space\vee\space x\approx\pm2\sqrt{2}\cdot\sqrt{\text{v}\cdot\text{t}}\tag4$$
Best Answer
Newton-Raphson is exactly the same for equations involving complex numbers. You just have to do the arithmetic using complex numbers.