سوال ۲۶ کنکور ارشد علوم کامپیوتر ۱۴۰۴

Suppose $z$ is a common root, so $z^{n}=1+i$ and $z^{m}=1-2i$ simultaneously.

Take moduli. Writing $r=|z|$, the modulus is multiplicative ($|z^{k}|=|z|^{k}$), so

$r^{n}=|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2},\qquad r^{m}=|1-2i|=\sqrt{1^{2}+2^{2}}=\sqrt{5}.$

Eliminate $r$. From the two equations $r=2^{1/(2n)}$ and $r=5^{1/(2m)}$, hence

$2^{1/(2n)}=5^{1/(2m)}.$

Raise both sides to the power $2mn$ (with $m,n\ge 1$):

$2^{m}=5^{n}.$

Use unique factorization. The left side has only the prime factor $2$ and the right side only the prime factor $5$. By the fundamental theorem of arithmetic, $2^{m}=5^{n}$ is impossible for any positive integers $m,n$ (it would force $m=n=0$, which are not natural numbers).

Therefore the moduli can never match, so the two equations share no common root for any natural $m,n$.

Exam trick: Before chasing arguments/angles, just compare moduli. A common root forces $|z|^{n}=|1+i|$ and $|z|^{n}=|1-2i|$ to be consistent; since $\sqrt{2}$ and $\sqrt{5}$ are powers of different primes, the system is dead on arrival.