solve 3-√-16÷1-√-9 in a+ib form
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Answer:
3-√-16÷1-√-9=3-√16×√-1 ÷1-√9×√-1
=3-√16 i ÷1-√9 i [As√-1=i]
=(3-4i )÷ (1-3i)
Rationalizing denominator
=(3-4i)÷(1-3i) ×(1-3i) ÷(1-3i)
On solving this, we get
=(3+9i-4i-12i²) ÷ 10
=(3+5i+12) ÷10 [As i²=(-1) ]
=(15+5i)÷10
=(3÷2)+(1÷2)i
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