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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Complex Number

NDA 2023 · Mathematics · Complex Number

What is the value of 12+5 i + 12-5 i , where i= -1 ?

What is the value of 12+5i+12−5i,wherei=−1?\sqrt{12+5 i}+\sqrt{12-5 i}, where i=\sqrt{-1} ?

Options

  1. A24
  2. B25
  3. C525 \sqrt{2}
  4. D5(2−1)5(\sqrt{2}-1)

Correct answer

C. 525 \sqrt{2}

Step-by-step solution

Let 12+5i=x+iy⇒12+5i=(x+iy)2⇒x2−y2=12\sqrt{12+5 \mathrm{i}}=\mathrm{x}+\mathrm{iy} \Rightarrow 12+5 \mathrm{i}=(\mathrm{x}+\mathrm{iy})^2\Rightarrow \mathrm{x}^2-\mathrm{y}^2=12...(i)and2xy=5and 2 x y=5...(ii) On solving (i) and (ii), we get x=52,y=12x=\frac{5}{\sqrt{2}}, \quad y=\frac{1}{\sqrt{2}} So, 12+5i=52+i2\sqrt{12+5 \mathrm{i}}=\frac{5}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}} Similarly, 12−5i=52−i2\sqrt{12-5 i}=\frac{5}{\sqrt{2}}-\frac{i}{\sqrt{2}} Now, 12+5i+12−5i=52+i2+52−i2=52\sqrt{12+5 i}+\sqrt{12-5 i}=\frac{5}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}+\frac{5}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}}=5 \sqrt{2}
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