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Answer 1

$\sf{αnswer:-}$

We have ( a+ib) ( a-ib) = a² + b². Also by data

a² + b² ≠ 0.

Thus, a ≠ 0 and b ≠ 0. That is a+ib ≠0.

By data, ( a+ib ) (x+ iy) = (a² + b²)i

⇒( a+ib ) (x+ iy) = ( a+ib ) = (a+ib)i

⇒x+iy = (a+ib)i ( ∵ a + ib ≠ 0 )

⇒x+iy = b + ai

⇒x = b and y = a

Answer 2

ANSWER:

Given:

• (a + bi)(x + iy) = (a² + b²)i, a,b ≠ 0

To Prove:

• y = a and x = b

Proof:

$\text{We are given that,}\\\\:\implies(a+bi)(x+iy)=(a^2+b^2)i\\\\\text{So,}\\\\:\implies(a+bi)(x+iy) = (a^2+b^2)i\\\\:\implies ax+bxi+ayi+byi^2= (a^2 + b^2)i\\\\\text{We know that, i^2=-1 }\\\\\text{So,}\\\\:\implies ax+by(-1)+(ay+bx)i=(a^2+b^2)i\\\\:\implies ax-by+(ay+bx)i=0+(a^2+b^2)i\\\\\text{Comparing real and imaginary parts,}\\\\\implies ax-by=0- - - -(1)\:\:\:\&\:\:\:ay+bx=a^2+b^2- - - -(2)\\\\\text{Solving (1),}\\\\:\implies ax-by=0\\\\:\implies ax=by\\\\:\implies x=\dfrac{by}{a}- - - -(3)$

$\text{Putting (3) in (2),}\\\\:\implies ay+bx=a^2+b^2\\\\:\implies ay+b\left(\dfrac{by}{a}\right)=a^2+b^2\\\\:\implies ay+\dfrac{b^2y}{a}=a^2+b^2\\\\:\implies \dfrac{a^2y+b^2y}{a}=a^2+b^2\\\\\text{On simplifying and cross-multiplying,}\\\\:\implies y(a^2+b^2)=a(a^2+b^2)\\\\:\implies y=a- - - -(4)\\\\\text{So,}\\\\:\implies x=\dfrac{by}{a}\\\\:\implies x=\dfrac{ba}{a} \\\\:\implies x=b- - - -(5)\\\\\text{So, from (4) \& (5)}\\\\\bf{:\implies y=a\:\:\:and\:\:\:x=b} \\\\\text{HENCE PROVED!!!}$