Question

Find the value of $\sqrt{(5+12i)}$, where $i=\sqrt{-1}$.

• (A) $\pm (2+3i)$
• (B) $\pm (3+2i)$
• (C) $\pm (2-3i)$
• (D) $\pm (1+2i)$

Answer 1

The Correct Option is B

Explanation:
Let, $\sqrt{(5+12i)}=x+iy$On squaring both sides we get,$5+12i=(x+iy{)}^2$$\Rightarrow 5+12i=x^2+(iy{)}^2+2(x)(iy)$$\Rightarrow 5+12i=x^2+i^2{y}^2+2xyi$As we know,$i^2=-1$$\Rightarrow 5+12i=x^2+(-1)y^2+2xyi$$\Rightarrow 5+12i=(x^2-y^2)+(2xy)i$Comparing real and imaginary parts on both sides, we get$x^2-y^2=5$ and $2xy=12$$\therefore xy=6$$\Rightarrow y=\frac{6}{x}$Now, $(x^2-y^2)=5$Putting value of $y$ in above equation, we get$x^2-{\left(\frac{6}{x}\right)}^2=5$$\Rightarrow x^2-\frac{36}{x^2}=5$$\Rightarrow \frac{x^4-36}{x^2}=5$$\Rightarrow x^4-36=5x^2$$\Rightarrow x^4-5x^2-36=0$$\Rightarrow x^4-9x^2+4x^2-36=0$$\Rightarrow x^2(x^2-9)+4(x^2-9)=0$$\Rightarrow x^2-9=0$ or $x^2+4=0$$\Rightarrow x^2=9$ or $x^2=-4$$\Rightarrow x=\pm 3$ (we know that $x^2$ is always greater than zero so, we neglect $x^2=-4$Now, $x^2-y^2=5$$\therefore 9-y^2=5$$\Rightarrow y^2=4$$\Rightarrow y=\pm 2$So, $\sqrt{(5+12i)}=\pm (3+2i)$Hence, the correct option is (B).