Let z 1 = 1 0 + 6 i and z 2 = 4 + 6 i . And suppose z is any complex number such that the arguement of z − z 2 z − z 1 = 4 π .
Find the value of ∣ z − 7 − 9 i ∣ .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
In this solution I am going to use a geometrical approach. In the following graph the point A represents the complex number z 2 , B represents z 1 , C the number 7 + 9 i , and D the number z
The slope of the segment A C is 1 , and the slope of the segment C B is − 1 . Therefore, the angle A C B is 9 0 o . It is also easy to verify that A C = C B = 3 2 . Since A D B = ar g ( z − z 1 ) − ar g ( z − z 2 ) = ar g z − z 2 z − z 1 = 4 5 o , then A D B = 2 1 A C B . Thus D belongs to the circle with center at C and radius 3 2 and therefore ∣ z − 7 − 9 i ∣ = D C = 3 2 ≈ 4 . 2 4 2 .
Note: We consider only the case when D is above the horizontal line passing through A and B , because if D is below that line then it easy to see that ar g z − z 2 z − z 1 = ar g ( z − z 1 ) − ar g ( z − z 2 ) < 0 , and therefore is never going to be equal to 4 5 o .
Same method as John Ashley Capellan in L a T e x .
Let z = a + b i and f ( z ) = z − z 2 z − z 1 . Then:
f ( z ) = a − 4 + ( b − 6 ) i a − 1 0 + ( b − 6 ) i = [ ( a − 4 ) + ( b − 6 ) i ] [ ( a − 4 ) − ( b − 6 ) i ] [ ( a − 1 0 ) + ( b − 6 ) i ] [ ( a − 4 ) − ( b − 6 ) i ] = ( a − 4 ) 2 + ( b − 6 ) 2 ( a − 1 0 ) ( a − 4 ) + ( b − 6 ) 2 − ( a − 1 0 ) ( b − 6 ) i + ( a − 4 ) ( b − 6 ) i = ( a − 4 ) 2 + ( b − 6 ) 2 a 2 − 1 4 a + 4 0 + b 2 − 1 2 b + 3 6 + ( b − 6 ) ( − a + 1 0 + a − 4 ) i = ( a − 4 ) 2 + ( b − 6 ) 2 ( a 2 − 1 4 a + b 2 − 1 2 b + 7 6 ) + 6 ( b − 6 ) i
For ar g ( f ( z ) ) = 4 π ⇒ ℜ ( f ( z ) ) = ℑ ( f ( z ) )
⇒ a 2 − 1 4 a + b 2 − 1 2 b + 7 6 a 2 − 1 4 a + b 2 − 1 2 b + 7 6 − 6 b + 3 6 a 2 − 1 4 a + b 2 − 1 8 b + 1 1 2 ( a − 7 ) 2 − 4 9 + ( b − 9 ) 2 − 8 1 + 1 1 2 ( a − 7 ) 2 + ( b − 9 ) 2 ⇒ ∣ z − 7 − 9 i ∣ 2 ∣ z − 7 − 9 i ∣ = 6 ( b − 6 ) = 0 = 0 = 0 = 1 8 = 1 8 = 1 8
Prerequisite :
ar
g
(
z
−
z
2
z
−
z
1
)
=
θ
Represents,
1) Major arc of a circle passing through
z
1
and
z
2
when
θ
<
2
π
2) Semicircle when
θ
=
2
π
3) Minor arc when
θ
>
2
π
Solution :
Let the centre of the circle be P
(
a
,
b
)
and the variable point be A
(
x
,
y
)
The points
z
1
and
z
2
lie on the circle.Thus they are equidistant from the centre. Hence, the centre lies on the perpendicular bisector of line joining them, which is
R
e
(
z
)
=
7
Hence,
a
=
7
Now, draw four lines joining the variable point & centre with z 1 and z 2 as shown. ( Will post the image later. )
We know that the angle formed by the arc at centre is twice the angle formed by the arc at the circumference.
∴
∠
z
1
P
z
2
=
2
∠
z
1
A
z
2
=
2
π
Thus the centre lies on the circle formed with
z
1
z
2
as diameter.
Radius of this circle
=
∣
4
+
6
ι
−
(
7
+
6
ι
)
∣
=
3
Distance z 1 P = ∣ A − P ∣ = ∣ z 1 − P ∣ = 3 2 + 3 2 = 3 2
Problem Loading...
Note Loading...
Set Loading...
By letting z = a + bi and using the givens above, [z - z(1)]/[z - z(2)] = [(a^2 + b^2 - 14a - 12b + 76) + i(6b - 36)]/[a^2 - 8a + b^2 - 12b + 52].
From here, arg([z - z(1)]/[z - z(2)]) = arctan([6b - 36]/[a^2 + b^2 - 14a - 12b + 76]) = pi/4, which implies that (a - 7)^2 + (b - 9)^2 = 18.
Hence, |z - 7 - 9i| = |(a - 7) + i(b - 9)| = sqrt[(a - 7)^2 + (b - 9)^2] = sqrt(18).