a,b,c,d∈R such that a2+b2=4 and c2+d2=2 and if (a+ib)2=(c+id)2(x+iy) then x2+y2=
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Solution
Given that (a+ib)2=(c+id)2(x+iy) Taking modulus both the sides |(a+ib)2|=|(c+id)2(x+iy)| [∵|z1z2|=|z1||z2|,|zn|=|z|n] ∴|a+ib|2=|c+id|2|x+iy| ⇒a2+b2=(c2+d2)√x2+y2 Given that a2+b2=4,c2+d2=2 ⇒4=2√x2+y2 ⇒x2+y2=4