Given that u=x+yi. If z=(1-i)/(u+6) is a pure imaginary number,show that y=x+6. State the values of y if x=1.?

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Answer 1 · 2017-11-03T16:24:41

See the explanation below

#u=x+iy#

Then,

#z=(1-i)/(u+6)=(1-i)/(x+iy+6)#

Multiply numerator and denominator by the conjugate of the denominator

#z=(1-i)/(x+iy+6)*(x+6-iy)/(x+6-iy)#

#=((x+6)-iy-i(x+6)+i^2y)/((x+6)^2-y^2i^2)#

#=(x+6-y)/((x+6)^2+y^2)+((-x-y-6)i)/((x+6)^2+y^2)#

If #z# is pure imaginary number, then the real part is #ℜ(z)=0#

#ℜ(z)=(x+6-y)/((x+6)^2+y^2)=0#

The denominator is #(x+6)^2+y^2>0#, #AA (x,y) in RR^2#

Therefore,

#x+6-y=0#

#=>#, #y=x+6#

If #x=1#

#(1+6)^2+y^2>0#

#y^2> -49#

#y=+-7i#