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 $x=1$

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

$y^2> -49$

$y=+-7i$