If `abs(((3-i)(k+i))/(2-i))=10^(1/2)`, then the value of `k` is?

The modulus of a complex number `z=a+ib` is

`|z|=sqrt(a^2+b^2)`

Apply the following

`|((3-i)(k+i))/(2-i)|=sqrt10`

`<=>`, `|((3-i)||(k+i))|/|(2-i)|=sqrt10`

`<=>`, `(sqrt(9+1)sqrt(k^2+1))/(sqrt(4+1))=sqrt(10)`

`=>`

`(sqrt10*sqrt(k^2+1))/sqrt5=sqrt10`

`sqrt(k^2+1)=sqrt5`

`k^2+1=5`

`k^2=4`

`k=+-2`

First, by expanding the numerator:
`(3-i)(k+i)=3k+3i-ik-i^2=3k+1-ik+3i`

`(3k+1-ik+3i)/(2-i)=((3k+1-ik+3i)(2+i))/((2-i)(2+i))=(6k+2-2ki+6i+3ki+i-i^2k+3i^2)/(4-i^2)=(6k+2-2ki+6i+3ki+i+k -3)/(4+1)=(7k-1+7i+ki)/5=(7k-1)/5+(7+k)/5 i`

`abs(a+bi)=sqrt(a^2+b^2)`

`sqrt(((7k-1)/5)^2+((7+k)/5)^2)=sqrt10`

`((7k-1)/5)^2+((7+k)/5)^2=10`

`(7k-1)^2+(7+k)^2=250`

`49^2-14k+1+k^2+14k+49=250`

`50k^2-200=0`

`50k^2=200`

`k^2=4`

`k=+-2`

`(3-i)(k+i)=3k+3i-ik-i^2=(3k+1)+i(3-k)`.

`:. {(3-i)(k+i)}/(2-i)={(3k+1)+i(3-k)}/(2-i)xx(2+i)/(2+i)`,

`=[{(3k+1)+i(3-k)}(2+i)]/{4-i^2}`,

`=1/5[{2(3k+1)-(3-k)}+i{2(3-k)+(3k+1)}]`,

`=1/5{(7k-1)+i(k+7)}`.

`:. |{(3-i)(k+i)}/(2-i)|=1/5sqrt{(7k-1)^2+(k+7)^2}`,

`=1/5sqrt(50k^2+50)`,

`=sqrt{2(k^2+1)}`.

Hence, by the given, `sqrt{2(k^2+1)}=sqrt10`.

`:. k^2+1=5 rArr k=+-2`, is the desired value!

N.B.:- The Solution submitted by Respected Narad T. Sir is the
easiest and best one.