Given |z| = 3
Show that 7 ≤ |z+6-8i| ≤ 13
|z+6-8i| ≤ |z|+|6-8i|
[Triangle law of inequality]
\(\le 3\sqrt { { 6 }^{ 2 }+({ -8) }^{ 2 } } \le 3+\sqrt { 36+64 } \le +\sqrt { 1w } \)
|z+6-8i| ≤3+10 ≤ 13 ............... (1)
Also ||z+6-8i| ≥ |x|-|6-8i| ≥ \(|3-\sqrt { { 6 }^{ 2 }+(-8)^{ 2 } } |\)
≥ \(|3-\sqrt { 36+84 } |\) ≥ |3-10| ≥ |-7| ≥ 7 ............... (2)
From (1) and (2) we get,
7 ≤ |z+6-8i| ≤ 13.

\(\left| \frac { 2+i }{ -1+2i } \right| =\frac { \left| 2+i \right| }{ \left| -1+2i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } }{ \sqrt { \left( -1 \right) ^{ 2 }+{ 2 }^{ 2 } } } =1\) \(\left( \because \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| ,{ z }_{ 2 }\neq 0 \right) \)

We have z = (2+3i)(1−i) = (2+3)+(3−2)i = 5+i
\(\Rightarrow\) \({ z }^{ -1 }=\frac { 1 }{ z } =\frac { 1 }{ 5+i } \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

We have = \(\frac { z+3 }{ z-5i } =\frac { 1+4i }{ 2 } \)
\(\Rightarrow\) 2(z + 3) = (1 + 4i) (z− 5i)
\(\Rightarrow\) 2z + 6 = (1 + 4i)z + 20−5i
\(\Rightarrow\) (2−1−4i)z = 20− 5i− 6 
\(\Rightarrow\) \(z=\frac { 14-5i }{ 1-4i } =\frac { \left( 14-5i \right) \left( 1+4i \right) }{ \left( 1-4i \right) \left( 1+4i \right) } =\frac { 34+51i }{ 17 } =2+3i\)

i¹⁹⁴⁷+ i¹⁹⁵⁰
i¹⁹⁴⁷+i¹⁹⁵⁰ = i¹⁹⁴⁴.i³+i¹⁹⁴⁸.i²
[∴ 1944 is a multiple of 4, or 1948 is also a multiple of 4]
= (i⁴)486.i².i¹+(i⁴)487.i² [i⁴ = 1]
= (1⁴⁸⁶)(-1) + (1)⁴⁸⁷(-1) [i²= -1]
= -i-1
= -(1- i)

2 +i2\(\sqrt { 3 } \)
Let 2+i2\(\sqrt { 3 } \)  = x + iy = r (cosθ + i sinθ)
r = modulus =\(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
=\(\\ \sqrt { { 2 }^{ 2 }+(2\sqrt { 3 } )^{ 2 } } \)
= \(\sqrt { 4+12 } =\sqrt { 16 } \) = 4
α = tan-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { 2\sqrt { 3 } }{ 2 } \right| \)
= \(tan^{ -1 }(\sqrt { 3 } )=\frac { \pi }{ 3 } \)
Since the complex number 2+i2 \(\sqrt { 3 } \) lies in the I quadrant, [x, y both +ve] its principal value θ = α = \(\frac { \pi }{ 3 } \)
∴ Its polar form is 2+i2\(\sqrt { 3 } \)
= 4\(\left[ cos\left( 2k\pi +\frac { \pi }{ 3 } \right) +isin\left( 2k\pi +\frac { \pi }{ 3 } \right) \right] ,k\in Z\).

\(\left[ Re\left( iz \right) \right] ^{ 2 }=3\)
iz = i(x + iy) = ix + i²y = ix - y = -y + ix
⇒ Re(iz) = -y
[Re(iz)]² = -y
⇒ (-y)² = 3
⇒ y² = 3
Hence, the Cartesian equation is y² = 3

Given z = x + iy
Consider \(\left| \frac { z-4i }{ z+4i } \right| =1\Rightarrow \left| \frac { x+iy-4i }{ x+iy+4i } \right| \)=1
⇒ \(\left| \frac { x+i(y-4) }{ x+i(y+4) } \right| \)
⇒ \(\frac { \sqrt { { x }^{ 2 }+(y-4)^{ 2 } } }{ \sqrt { { x }^{ 2 }+((y+4)^{ 2 } } } \) = 1
⇒ \(\sqrt { { x }^{ 2 }+(y-4)^{ 2 } } =\sqrt { { x }^{ 2 }+(y+4)^{ 2 } } \)
Squaring both sides we get,
x²+(y-4)² = x²+(y+4)²
\(\Rightarrow \not x^{2}+\not y^{2}-8 y+\not 16=\not x^{2}+\not y^{2}+8 y+\not 16\)
⇒ 8y+8y = 0
⇒ 16y = 0
⇒ y = 0    [∵ 16 ≠ 0]
y = 0 is the equation of real axis Locus of z is the real axis.

It is enough to prove that the sides of the triangle are equal.
Let z₁ = 1, \({ z }_{ 2 }=\frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \) and \({ z }_{ 3 }=\frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)
The length of the sides of the triangles are
\(\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\left| 1-\left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\left| \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\frac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } \)
\(\left\lfloor { z }_{ 2 }-{ z }_{ 3 } \right\rfloor =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -\left( \frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 } } =\sqrt { 3 } \)
\(\left| { z }_{ 3 }-{ z }_{ 1 } \right| =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -1 \right| =\left| \frac { -3 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\sqrt { 3 } \)
Since the sides are equal, the given points form an equilateral triangle

We consider \(\frac { 1+i }{ 1-i } =\frac { \left( 1+i \right) \left( 1+i \right) }{ \left( 1-i \right) \left( 1+i \right) } =\frac { 1+2i }{ 1+1 } =\frac { 2i }{ 2 } =i\)
and \(\frac { 1-i }{ 1+i } =\left( \frac { 1+{ i } }{ 1-i } \right) ^{ -1 }=\frac { 1 }{ i } =-i\)
Therefore,\(\left( \frac { 1+i }{ 1-i } \right) ^{ 3 }-\left( \frac { 1-i }{ 1-i } \right) ^{ 2 }\)= i³-(-i)³ = - i - i = -2i

 \(\left( -\sqrt { 3 } +3i \right) ^{ 31 }\)
Let \(-\sqrt { 3 } +3i=r\left( cos\theta +isin\theta \right) \). Then, we get
\(r=\sqrt { \left( -\sqrt { 3 } \right) ^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 12 } =2\sqrt { 3 } \)
\(\alpha ={ tan }^{ -1 }\left| \frac { 3 }{ -\sqrt { 3 } } \right| ={ tan }^{ -1 }\sqrt { 3 } =\frac { \pi }{ 3 } \)
\(\theta =\pi -\alpha =\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \) (\(\because\)  \(\sqrt { 3 } +3i\) lies in II Quadrant)
Therefore,\(-\sqrt { 3 } +3i=2\sqrt { 3 } \left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
Raising power 31 on both sides,
\(\left( -\sqrt { 3 } +3i \right) ^{ 31 }=\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) ^{ 31 }\)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( 20\pi +\frac { 2\pi }{ 3 } \right) +isin\left( 20\pi +\frac { 2\pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( \pi -\frac { \pi }{ 3 } \right) +isin\left( \pi -\frac { \pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) =\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \).

We have to find \((\sqrt{3}+1)^{\frac{1}{3}}\). Let \(z=(\sqrt{3}+i)^{\frac{1}{3}}\). Then \({ z }^{ 3 }=\sqrt { 3 } +i=r\left( cos\theta +isin\theta \right) \)
Then, \(r=\sqrt { 3+1 } =2\) and \(\alpha =\theta =\frac { \pi }{ 6 } \) (\(\because \sqrt{3}+i\) lies in the first quadrant)
Therefore, \({ z }^{ 3 }=\sqrt { 3 } +i=2\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow z=\sqrt [ 3 ]{ 2 } \left( cos\left( \frac { \pi +12k\pi }{ 18 } \right) +isin\left( \frac { \pi +12k\pi }{ 18 } \right) \right) \), k = 0, 1, 2.
Taking k = 0, 1, 2, we get
k = 0, z \(={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 1, \(z={ z }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 2, \(z={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { 25\pi }{ 18 } +sin\frac { 25\pi }{ 18 } \right) ={ 2 }^{ \frac { 1 }{ 3 } }\left( -cos\frac { 7\pi }{ 18 } -sin\frac { 7\pi }{ 18 } \right) \)

z³ = -27 = (-1 \(\times\) 3)³ = -1 \(\times\) 33
z = \((-1)^{ \frac { 1 }{ 3 } }\times 3^{ 3\times \frac { 1 }{ 3 } }=(-1)^{ \frac { 1 }{ 3 } }\)\(\times\) 3
∴ z = 3\(\left[ cos\pi +isin\pi \right] ^{ \frac { 1 }{ 3 } }\)
[∵ cos π = -1 and sin π = 0]
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \)
k = 0, 1, 2
When k = 0,
z = 3\(\left[ cos\frac { 1 }{ 3 } (\pi )isin\frac { 1 }{ 3 } (\pi ) \right] =3cos\frac { \pi }{ 3 } \)
When k = 1
z = 3\(\left[ cos\frac { 1 }{ 3 } (3\pi )isin\frac { 1 }{ 3 } (3\pi ) \right] \)
= 3[cos π + i sin π] = 3(-1+0) 
When k = 2
z = 3\(\left[ cos\frac { 1 }{ 3 } (5\pi )isin\frac { 1 }{ 3 } (5\pi ) \right] =3\left[ cos5\frac { \pi }{ 3 } \right] \)
Hence, the roots are 3 cis\(\frac { \pi }{ 3 } \), -3, 3 c is 5\(\frac { \pi }{ 3 } \)

Given z = x + iy
Im \(\left( \frac { 2z+1 }{ iz+1 } \right) \)= 0
⇒ Im\(\left( \frac { 2(x+iy)+1 }{ i(x+iy)+1 } \right) \)= 0
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix+i^{ 2 }y+1 } \right) \)
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix-y+1 } \right) \)
\(\left( \frac { (2x+1)+iy }{ (1-y)+ix } \right) \)
Multiply and divide by the conjugate of the denominator
We get Im\(\left( \frac { (2x+1)+2iy }{ (1-y)+ix } \times \frac { (1-y)-ix }{ (1-y)-ix } \right) \)=0
⇒ Im\(\left( \frac { (2x+1)+2iy\times (1-y)-ix }{ (1-y)^{ 2 }+{ x }^{ 2 } } \right) \)
Choosing the imaginably part we get,
\(\frac { (2x+1)(-x)+2y(1-y) }{ (1-y)^{ 2 }+{ x }^{ 2 } } \)
⇒ (2x+1)-x+2y(1-y) = 0
⇒ -2x²-x+2y-2y² = 0
⇒ 2x²+2y²+x-2y = 0
Hence, locus of z is 2x²+2y²+x-2y = 0

Comparing the two given lines with
\(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have, \(\vec { a } =-\hat { -1 } -3\hat { j } -5\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +7\hat { k } ,\vec { c } =2\hat { i } +4\hat { j } +6\hat { k } \) and \(\vec { d } =\hat { i } +4\hat { j } +7\hat { k } \)
We know that the two given lines are coplar, if \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)=0
Here, \(\vec { b } \times \vec { d } \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| =7\hat { i } -14\hat { j } +7\hat { k } \) and \(\vec { c } -\vec { a } =3\hat { i } +7\hat { j } +11\hat { k } \)
Then, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(3\hat { i } +7\hat { j } +11\hat { k } )(7\hat { i } -14\hat { j } +7\hat { k } )\)
Therefore the two given lines are coplanar.Then we find the non parametric form of vector equation of the plane containing the two given coplanar lines. We know that the plane containing the two given coplanar lines is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { d } )\)=0
which implies that \((\vec { r } -(-\hat { i } -3\hat { j } -5\hat { k } )).(7\hat { i } -14\hat { j } +7\hat { k } )\)=0. Thus, the required non-parametric vector equation of the plane containing the two given coplanar lines is \(\vec { r } .(\hat { i } -2\hat { j } +\hat { k } )\)=0.

-8

x-iy