= Solution
Choose a nonzero <integral ideal> $B$ whose class is $[A]^{-1}$; one can multiply the fractional inverse $A^{-1}$ by a suitable nonzero <integer> to obtain such a $B$. By the given bound choose $0\ne\beta\in B$ with $|N(\beta)|\le c_KN(B)$. Put $I_A=(\beta)B^{-1}$. Since $(\beta)\subseteq B$, multiplication by $B^{-1}$ makes $I_A$ an <integral ideal>. Its class is $[B]^{-1}=[A]$, and multiplicativity of <ideal norms> gives
$$
\boxed{N(I_A)=\frac{|N(\beta)|}{N(B)}\le c_K.}
$$
This proves the class representative bound without accidentally constructing a representative of the inverse class instead.
To determine the <Ideal class group of Q of square root minus thirty-four>, write $\eta=\sqrt{-34}$. The given <integral basis> has <discriminant> $-136$, so the imaginary-quadratic bound is $4\sqrt{34}/\pi<8$. Every class therefore has a representative of <ideal norm> at most seven, and factoring such representatives shows that the <ideal class group> is generated by <prime ideals> of <ideal norm> at most seven.
Two ramifies: let $P_2=(2,\eta)$, with $P_2^2=(2)$. Three is inert because $X^2+34\equiv X^2+1$ is irreducible modulo three, so its <prime ideal> <ideal norm> is nine and supplies no small generator. At five and seven the roots are $\pm1$, so the primes of <ideal norm> five and seven are conjugate pairs. Set $P_5=(5,\eta-1)$ and $P_7=(7,\eta+1)$. Their conjugates have inverse <ideal> classes. No new <prime ideal> has <ideal norm> four, since the only prime above two has <residue degree> one; six is not a <prime power>. Thus $[P_2],[P_5],[P_7]$ generate the <ideal class group>.
The element <field norms> $N(4+\eta)=50$ and $N(1+\eta)=35$, together with reduction at the split primes, give
$$
(4+\eta)=P_2P_5^2,\qquad(1+\eta)=\bar P_5P_7.
$$
The <ideal norm> products verify both exponents. Consequently $[P_5]^2=[P_2]$ and $[P_7]=[P_5]$, so the whole <group> is generated by $[P_5]$ and its order divides four. The class $[P_2]$ is not trivial: a generator would be an element of the <ring of integers> $a+b\eta$ of <field norm> two, but $a^2+34b^2=2$ has no <integer> solution. Since $[P_2]^2=1$, its order is exactly two. Hence $[P_5]$ has order four and
$$
\boxed{\operatorname{Cl}(\mathbb Q(\sqrt{-34}))\cong C_4.}
$$
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