Question:
If \(\sqrt{-24-10 i}=\alpha+i \beta\), \(i=\sqrt{-1}\), \(\alpha, \beta \in \mathbf{R}\), then \(\alpha^{2}+\beta^{2}\) is equal to :
If \(\sqrt{-24-10 i}=\alpha+i \beta\), \(i=\sqrt{-1}\), \(\alpha, \beta \in \mathbf{R}\), then \(\alpha^{2}+\beta^{2}\) is equal to :
Show Hint
Take the modulus on both sides: \(|\alpha+i\beta|^2=|-24-10i|\).
Updated On: Nov 12, 2026
- \(22\)
- \(24\)
- \(26\)
- \(29\)
Show Solution
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The Correct Option is C
Solution and Explanation
Square both sides: \((\alpha+i\beta)^2=-24-10i\), i.e. \(\alpha^2-\beta^2=-24\) and \(2\alpha\beta=-10\).
Taking moduli: \(|\alpha+i\beta|^2=|-24-10i|\).
\(\alpha^2+\beta^2=\sqrt{(-24)^2+(-10)^2}=\sqrt{576+100}=\sqrt{676}=26\).
Answer: (3) \(26\)
Taking moduli: \(|\alpha+i\beta|^2=|-24-10i|\).
\(\alpha^2+\beta^2=\sqrt{(-24)^2+(-10)^2}=\sqrt{576+100}=\sqrt{676}=26\).
Answer: (3) \(26\)
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