Question

The square roots of the complex number $(-5 - 12i)$ are

• A. $\pm (2 - 3i)$
• B. $\pm (3 + 2i)$
• C. $\pm (2 + 3i)$
• D. $\pm (3 - 2i)$

Hint:

Bypass long algebraic expansions entirely during multiple-choice tests by squaring the options! For option (A): $(\pm(2-3i))^2 = 4 + 9i^2 - 12i = 4 - 9 - 12i = -5 - 12i$. Since this squares back to our original number instantly, it confirms the answer in seconds.

Answer 1

The Correct Option is A

Step 1: Use the polar shortcut for square roots.
For $z=-5-12i$, the modulus is $|z|=\sqrt{(-5)^2+(-12)^2}=\sqrt{25+144}=13$.
Step 2: Apply the real-part formula.
If $\sqrt z=a+ib$, then $a^2-b^2=\Re(z)=-5$ and $a^2+b^2=|z|=13$.
Step 3: Solve for $a^2$ and $b^2$.
Adding: $2a^2=8\Rightarrow a^2=4\Rightarrow a=\pm2$. Subtracting: $2b^2=18\Rightarrow b^2=9\Rightarrow b=\pm3$.
Step 4: Fix the sign using the imaginary part.
Since $2ab=\Im(z)=-12<0$, the product $ab$ must be negative, so $a$ and $b$ have opposite signs.
Step 5: Pick the matching pair.
Take $a=2,\,b=-3$ (or $a=-2,\,b=3$), giving the root $2-3i$ and its negative.
Step 6: Write both roots.
Therefore the square roots are $\pm(2-3i)$. \[ \boxed{\pm(2-3i)} \]