Question

let z1=10+6i, z2=4+6i,if z is any complex number such that the argument z-z1/z-z2=pie/4 then the value of |z-7-9i| is?

Answer 1

@Sambhavvinaykya

Answer 2

@Nnesha

Answer 3

take conjugation first and separate the real and imaginary part

Answer 4

can u solve whole of it.difficult to solve

Answer 5

hmm ok after take conjugation and you get
\[\frac{ ((x-10)+i(y-6))((x-4)-i(y-6)) }{(x-4)^{2}+(y-6)^{2}}\]

Answer 6

confused.

Answer 7

@danish071996 - what part are you confused on?

Answer 8

I believe what @Sambhavvinaykya did was to first let:\[z=x+iy\]and then substitue this into:\[\frac{z-z_1}{z-z_2}\]does that make sense so far?

Answer 9

@danish071996 - if you don't answer then it becomes very difficult to help

Answer 10

yes i can understand that

Answer 11

ok, so if do the substitution we get:\[\frac{x+iy-(10+6i)}{x+iy-(4+6i)}=\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)}\]agreed?

Answer 12

yes.after that

Answer 13

what he did next was to multiply the numerator and the denominator by the complex conjugate of the denominator

Answer 14

that would leave you with a non-complex number in the denominator - understand?

Answer 15

yes.bt how to find the value?

Answer 16

so, if we follow through we get:\[\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)}=\frac{ ((x-10)+i(y-6))((x-4)-i(y-6)) }{(x-4)^{2}+(y-6)^{2}}\]\[=\frac{x^2-14x+y^2-12y+76+6(y-6)i}{(x-4)^2+(y-6)^2}\]

Answer 17

your question states that the argument of this complex number is \(\pi/4\) - do you know what an argument of a complex number is?

Answer 18

yes i know

Answer 19

good, so you know that if we had some complex number, say \(a+ib\), and we were told that its argument was \(\pi/4\), then this implies:\[\frac{\pi}{4}=\tan^{-1}(\frac{b}{a})\]agreed?

Answer 20

yes.agree

Answer 21

good, this also therefore implies that:\[\tan(\frac{\pi}{4})=\frac{b}{a}=1\]\[\therefore b=a\]agreed?

Answer 22

yes

Answer 23

good, now if we look back at the result we got:\[\frac{x^2-14x+y^2-12y+76+6(y-6)i}{(x-4)^2+(y-6)^2}\]this implies that the real and imaginary components are equal. Also, since the denominator is a real number, we can just concentrate on the numerator and deduce that:\[x^2-14x+y^2-12y+76=6(y-6)\]agreed?

Answer 24

yes

Answer 25

ok, so now lets rearrange this a little:\[x^2-14x+y^2-12y+76=6y-36\]\[\therefore x^2-14x+y^2-18y+112=0\]\[(x-7)^2+(y-9)^2-18=0\]\[\therefore (x-7)^2+(y-9)^2=18\]agreed?

Answer 26

yes.

Answer 27

now we are almost there, the question is asking you to find:\[|z-7-9i|\]if we substitute \(z=x+iy\) into this we get:\[|z-7-9i|=|x+iy-7-9i|=|(x-7)+i(y-9)|=?\]hopefully you can complete this from here?

Answer 28

yes.i got it.thanks

Answer 29

yw :)

Answer 30

sorry internet connectivity got lost so couldnt reply @danish071996
nice explanation @asnaeer

Answer 31

@danish071996