# Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.6

Question 1.
If 2 = x + iy is a complex number such that $$\left|\frac{z-4 i}{z+4 i}\right|$$ = 1 show that the locus of z is real axis.
Solution:
Let z = x + iy
$$\left|\frac{z-4 i}{z+4 i}\right|$$ = 1 ⇒ |z – 4i| = |z + 4i|
⇒ |x + iy – 4i| = |x + iy + 4i|
⇒ |x + (y – 4)i| = |x + (y + 4)i|
x² + (y – 4)² = x² + (y + 4)²
⇒ (y – 4)² = (y + 4)² = ⇒ y – 4 = ±(y + 4)
y – 4 = -y – 4
2y – 0; y – 0 ⇒ x-axis
The locus is real axis

Samacheer Kalvi 12th Maths Guide

Question 2.
If z = x + iy is a complex number such that Im $$\left(\frac{2 z+1}{i z+1}\right)$$ = 0, show that the locus of z is 2x² + 2y² + x – 2y = 0.
Solution:
Let z = x + iy

⇒ 2y – 2y² – 2x² – x = 0
⇒ 2x² + 2y² – 2y + x = 0

Question 3.
Obtain the Cartesian form of the locus of z = x + iy in each of the following cases:
(i) [Re (iz)]² = 3
Solution:
since x + iy = z
iz = ix – y
∴ Re (iz) = -y
⇒ (y)² = 3
⇒ y² – 3 = 0
y² = 3

(ii) Im[(1 – i)z + 1] = 0
Solution:
Im ((1 – i) (x + iy) + 1) = 0
Im [x – ix + iy + y + 1] = 0
⇒ -x + y = 0 or x – y = 0

(iii) |z + i| = |z – 1|
Solution:
|x + iy + i| = |x + (y – 1)|
|x + i(y + 1)| = |x – 1) + iy
⇒ x² + (y + 1)² = (x – 1)² + y²
x² + y² + 2y + 1 = x² – 2x + 1 + y²
⇒ 2x + 2y = 0
x + y = 0

(iv) $$\bar {z}$$ = z-1
Solution:

x² + y² = 1, x² + y² = -1 which cannot be true.
∴ x² + y² = 1

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) |z – 2 – i| = 3
Solution:
Let z = x + iy
|z – 2 – i| = 3 ⇒ |z – (2 + i)| = 3
∴ It is a circle with centre at 2 + i
i.e., (2, 1) and radius = 3 units

(ii) |2z + 2 – 4i| = 2
Solution:
2 |z + 1 – 2i| = 2
⇒|z – (-1 + 2i)| = 1
⇒ It is a circle with centre at (-1 + 2i)
⇒ (-1, 2) and radius = 1 unit

(iii) |3z – 6 + 12i| = 8.
Solution:
3 |z – 2 + 4i| = 8
⇒ |z – (2 – 4i)| = $$\frac{8}{3}$$
It is a circle with centre at (2 – 4i) ⇒ (2, -4) and radius $$\frac{8}{3}$$ units.

Question 5.
Obtain the Cartesian equation for the locus of z = x + iy in each of the following cases:
(i) |z – 4| = 16
Solution:
Let z = x + iy
|x + iy – 4| – 16
|(x – 4) + iy| = 16
(x – 4)² + y² = 16²
∴ It is a circle with centre at (4, 0) and radius 16
x² + y² – 8x + 16 = 256
x² + y² – 8x – 240 = 0

(ii) |z – 4|² – |z – 1|² = 16.
Solution:
|x + iy – 4|² – |x + iy – 1|² = 16
|(x – 4) + iy|² – |(x – 1) + iy|² = 16
(x-4)² + y² – [(x – 1)² + y²] = 16
x² – 8x + 16 + y² – x² + 2x – 1 – y² = 16
-6x – 1 = 0 ⇒ 6x + 1 = 0

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