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.