سوال ۲۶ کنکور ارشد مهندسی کامپیوتر ۱۴۰۳

Write $1+i$ in polar form. Its modulus is $|1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$ and its argument is $\frac{\pi}{4}$, so as a multivalued expression

$1+i = \sqrt{2}\,e^{i\left(\frac{\pi}{4} + 2k\pi\right)}, \quad k \in \mathbb{Z}.$

Modulus of the result ($x$). Raising to the power $\frac{4}{3}$ raises the modulus to the same power:

$x = \left(\sqrt{2}\right)^{4/3} = \left(2^{1/2}\right)^{4/3} = 2^{2/3} = \sqrt[3]{4}.$

This already eliminates options 1 and 2, which claim $x = 2\sqrt{2}$.

Argument of the result ($y$). Multiply the argument by $\frac{4}{3}$:

$y = \frac{4}{3}\left(\frac{\pi}{4} + 2k\pi\right) = \frac{\pi}{3} + \frac{8k\pi}{3} = \frac{(8k+1)\pi}{3}.$

Because the exponent $\frac{4}{3}$ has denominator $3$, the expression $(1+i)^{4/3}$ has exactly three distinct values. Taking $k = 0, 1, 2$ and reducing modulo $2\pi$:

$\frac{\pi}{3}, \qquad 3\pi \equiv \pi = \frac{3\pi}{3}, \qquad \frac{17\pi}{3} \equiv \frac{5\pi}{3}.$

So the three arguments are $\dfrac{\pi}{3}, \dfrac{3\pi}{3}, \dfrac{5\pi}{3}$, which is exactly the family

$y = \frac{(2k+1)\pi}{3}, \quad k \in \mathbb{Z}.$

This matches option 3.

Pitfall: Option 4 only gives the principal branch $\frac{\pi}{3} + 2k\pi$ and misses two of the three roots of the fractional power. When the exponent is a fraction $p/q$, always include the $+2k\pi$ ambiguity of the argument before multiplying, then collect all $q$ distinct branches.