The square root of $7+24 i$
The square root of $7+24 i$
- $4-3 i$
- $3+4 i$
- $3-4 i$
- $4+3 i$
Solution
Let $\sqrt{7+24 i}=x+i y \Rightarrow 7+24 i=\left(x^2-y^2\right)+2 x y i$
Equating real and imaginary parts
$\begin{aligned}
& x^2-y^2=7 \text { and } 2 x y=24 \Rightarrow y=\frac{12}{x} \\
& \therefore x^2-\left(\frac{12}{x}\right)^2=7 \Rightarrow x^4-7 x^2-144=0 \\
& \Rightarrow x^2=16,-9 \Rightarrow x= \pm 4, y= \pm 3
\end{aligned}$
So, $(x+i y)= \pm(4+3 i)$.
Asked in: AP EAMCET 2024 (21 May Shift 2)
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