Question

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Answer 1

Answer:

If  $\frac{a+3i }{2+ib} =1-i$  , show that $(5a-7b)=0$

$a+3i=(2+ib)(1-i)$

$a+3i=2-2i+ib-bi^2$

$i^{2} = -1$ ⇒ $bi^2 = -b$

$a+3i=2-2i+ib-\left(-b\right)$

$a+3i=2-2i+ib+b$

$a+3i=(2+b)+i(b-2)$

$a = 2+b\\a-b=2\\a=7$                           $3=b-2\\b=5$

Prove that:

$(5a-7b)=0$

$(5*7-7*5)=0\\(35-35)=0\\0=0$

Done.