Question

Please HELP !!!: Complex Number

Answer 1

The complex number u is defined by :
\[u = \frac{ 6-3i }{ 1+2i }\]

Answer 2

For complex numbers z satisfying arg ( z -u) = Pi/4 , find the least possible value of |z|

Answer 3

@mathslover

Answer 4

First simplify u.

Answer 5

ok then ?

Answer 6

What did you get for u ?

Answer 7

-3i ...

Answer 8

No, no, it will be like this :

\(\cfrac{6-3i}{1-2i}\)
\(\cfrac{(6-3i)(1-2i)}{(1+2i)(1-2i)}\)
\(\cfrac{6-12i-3i-6}{5}\)
\(\cfrac{-15i}{5} = -3i \)

Yeah, right now :)

Answer 9

Now see , we have : arg(z-u) = arg(z-(-3i)) = arg(z+3i)

right?

Answer 10

arg ( z + 3i ) = Pi/4

Answer 11

Yeah, right, now we have : arg(z+3i) = \(\tan^{-1}(1)\)

any problem in this? ^

Answer 12

@mathsolver-i think for simplification u should multiply with conjugate

Answer 13

@08surya yep, right but I think I have done it correct. We had denominator as 1+2i , its conjugate will be 1-2i, so multiplied both numerator and denominator with 1-2i ...

Answer 14

then ?

Answer 15

can you tell me what is : arg(a+bi) ?

Answer 16

@mathslover -actuallly i was talking about ur simplification where u written denominator=1-2i

Answer 17

k , sorry for that, typing mistake, thanks for pointing it out.

Answer 18

tan inverse.. (b/a )

Answer 19

its all right

Answer 20

Good @antoni7

Now : let z = x+ iy

you get : arg(z+3i) = arg(x+i(y+3)) => arg(x+i(y+3)) = \(\tan^{-1}{(1)}\)

or : \(\tan^{-1}{(\cfrac{y+3}{x})} = \tan^{-1}{(1)}\)

Answer 21

ok...

Answer 22

Now you get : (y+3)/x = 1

=> y+3 = x
=> x^2 = (y+3)^2

Since : \(|z| = \sqrt{x^2 + y^2}\)

=> \(|z|^2 = x^2 + y^2\)

=> \(|z|^2 = (y+3)^2 + y^2 \) ( as : \(x^2 = (y+3)^2\) )

Answer 23

what is the least possible value of |z| :S ?

Answer 24

Differentiating : \(\cfrac{d((y+3)^2 + y^2)}{dy} = 0 \)

Solve for y now.

Answer 25

you're very very near to the answer antoni, have patience! Please.

Answer 26

wait wait wait ....

Answer 27

y = -1 ?

Answer 28

No... check your soln again