Skip to content
Exercise 1.2 · Q63

Q.Find the square root of the following complex number : 3+210 i3+2\sqrt{10}\,i

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
30% · 63/208 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Let 3+210 i=a+ib\sqrt{3+2\sqrt{10}\,i}=a+ib. Then a2−b2=3a^2-b^2=3, 2ab=2sqrt102ab=2\\sqrt{10}. (a2+b2)2=32+(2sqrt10)2=9+40=49Rightarrowa2+b2=7(a^2+b^2)^2=3^2+(2\\sqrt{10})^2=9+40=49\\Rightarrow a^2+b^2=7. Adding with a2−b2=3a^2-b^2=3: 2a2=10Rightarrowa2=5Rightarrowa=pmsqrt52a^2=10\\Rightarrow a^2=5\\Rightarrow a=\\pm\\sqrt5; subtracting: $2b^2=4\Rightarrow b^2=2\Rightarrow b=\pm\ …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.