Question

Find the real number a and b?

`(a+2i)(2-bi)=7-i`

Answer 1

We do not have `a,b` as real numbers, either `a=(7-isqrt31)/4` and `b=(7+isqrt31)/4` or `a=(7+isqrt31)/4` and `b=(7-isqrt31)/4`

Explanation:

`(a+2i)(2-bi)=7-i`

`=>2a-abi+4i-2bi^2=7-i`

`=>2a-abi+4i+2b=7-i`

`=>(2a+2b)+(4-ab)i=7-i`

Hence `2a+2b=7` i.e. `a+b=7/2` or `a=7/2-b`

and `4-ab=-1` and putting `a=7/2-b` in this we get

`4-(7/2-b)b=-1`

or `8-7b+2b^2=-2`

i.e. `2b^2-7b+10=0`

As discriminant is `7^2-4*2*10=-31`, we do not have real `a` and `b`.

`b=(7+-sqrt(49-80))/4=(7+-isqrt31)/4`

and if `b=(7+isqrt31)/4`

`a=7/2-(7+isqrt31)/4=(7-isqrt31)/4`

and if `b=(7-isqrt31)/4`

`a=7/2-(7-isqrt31)/4=(7+isqrt31)/4`

Hence either `a=(7-isqrt31)/4` and `b=(7+isqrt31)/4`

or `a=(7+isqrt31)/4` and `b=(7-isqrt31)/4`