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Exercise 1.2 · Q61

Q.Find the square root of the following complex number : 7+24i7+24i

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✓ Free question

Let 7+24i=a+ib\sqrt{7+24i}=a+ib. Then a2−b2=7a^2-b^2=7, 2ab=242ab=24. (a2+b2)2=72+242=49+576=625⇒a2+b2=25(a^2+b^2)^2=7^2+24^2=49+576=625\Rightarrow a^2+b^2=25. Adding with a2−b2=7a^2-b^2=7: 2a2=32⇒a2=16⇒a=±42a^2=32\Rightarrow a^2=16\Rightarrow a=\pm4; subtracting: 2b2=18⇒b2=9⇒b=±32b^2=18\Rightarrow b^2=9\Rightarrow b=\pm3. Since 2ab=24>02ab=24>0, a,ba,b share the same sign: (4,3)(4,3) or (−4,−3)(-4,-3).

✓Final answer

7+24i=±(4+3i)\sqrt{7+24i}=\pm(4+3i).

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