If #abs(((3-i)(k+i))/(2-i))=10^(1/2)#, then the value of #k# is?
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"How does implicit differentiation work?"
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.
