<s>[INST] Two unrelated equations, $w^2 = -\frac{15}{4} - 2i$ and $z^2 - (3-4i)z + (2-4i) = 0$? I am to solve the equation $z^2 - (3-4i)z + (2-4i) = 0$, and also have to help me that $w^2 = -\frac{15}{4} - 2i$. 
I can ofcourse find $w$, but I fail to see how this helps me in solving $z^2 - (3-4i)z + (2-4i) = 0$. What am I missing? Is there a connection I fail to see?
 [/INST] $z=\frac{3-4i±\sqrt{(3-4i)^2-4\cdot 1\cdot (2-4i)}}{2}$
Now, $(3-4i)^2-4\cdot 1\cdot (2-4i)=-15-16i=1^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
So, $z=\frac{3-4i±(1-4i)}{2}=1$ or $2-4i$
$4w^2=-15-8i=(1)^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
$2w=±(1-4i)$

Alternatively, let $-\frac{15}{4}-2i=(a-ib)^2$
So, $-\frac{15}{4}-2i=a^2-b^2-i2ab$
$a^2-b^2=-\frac{15}{4}$ and $2ab=2$
$(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2=(a^2-b^2)^2+(2ab)^2=\frac{289}{16}$
$a^2+b^2=\frac{17}{4}$
$a^2=\frac{1}{4},b^2=4$
So,$a=±\frac{1}{2},b=±2$ as $a,b$ are of same sign.
The square root of $(3-4i)^2-4\cdot 1\cdot (2-4i)=-15-16i$ can be calculated in the same way.
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