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Miscellaneous Exercise 1 (Subjective) · Q195

Q.Show that: [the printed source's nested-radical layout is corrupted at this item — could not reliably reconstruct the verbatim stem]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
94% · 195/208 Questions

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Concept understanding — Square Root of a Complex Number

To find the square root of a complex number x+iyx+iy, assume the (unknown) square root has the form a+iba+ib for real a,ba,b, so x+iy=a+ib\sqrt{x+iy}=a+ib. Squaring both sides gives x+iy=(a+ib)2=(a2−b2)+i(2ab)x+iy=(a+ib)^2=(a^2-b^2)+i(2ab), and equating real and imaginary parts on the two sides produces a pair of simultaneous real equations, x=a2−b2x=a^2-b^2 and y=2aby=2ab, in the two unknowns aa and bb. These are solved together — typically by using the auxiliary identity (a2+b2)2=(a2−b2)2+(2ab)2=x2+y2(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=x^2+y^2, which gives a2+b2a^2+b^2 directly as x2+y2\sqrt{x^2+y^2}; combining this with a2−b2=xa^2-b^2=x then pins down a2a^2 and b2b^2 separately by simple addition and subtraction, and taking square roots gives aa and bb up to sign. The signs of aa and bb are not independent: because y=2aby=2ab fixes the relative sign of aa and bb (same si …

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