Number Theory Test 20

$$\dfrac{(\omega - \overline{\omega} z)}{1-z}$$ is purely real and $$z\ne 1$$

$$\left|z_1 + z_2\right| \le \left|z_1\right| + \left|z_2\right| = \left|24 + 7i\right| + 6 = 25 + 6 = 31$$
Also, 
$$\left|z_1 + z_2\right| = \left|z_1 - (-z_2)\right| \ge \left|\left|z_1\right| - \left|z_2\right|\right|
\Longrightarrow \left|z_1 + z_2\right| \ge \left|25 - 6\right| = 19$$
Hence, the least value of $$\left|z_1 + z_2\right|$$ is 19 and the greatest value is 31.

Ans: C 

Given that $$(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$$
Taking modulus and squaring both sides, we get 
$$(8 + 1)^{50} = 3^{98}(a^2 + b^2)$$
or $$9^{50}=3^{98}(a^2 + b^2)$$
or $$3^{100} = 3^{98} (a^2 + b^2)$$
or $$(a^2 + b^2) = 9$$

Ans: A

$$\left| z-\frac { 4 }{ z }  \right| =2$$

$$\Longrightarrow \left| { z }^{ 2 }-4 \right| =2\left| z \right| $$       ..(1)

$$z=rcisA=r\left( \cos { A } +i\sin { A }  \right) $$

Substitute $$z$$ in equation (1)
$$\Longrightarrow \left| { r }^{ 2 }\cos { 2A-4+{ r }^{ 2 }i\sin { 2A }  }  \right| =2r$$

$$\Longrightarrow { \left( { r }^{ 2 }\cos { 2A } -4 \right)  }^{ 2 }+{ r }^{ 4 }\sin ^{ 2 }{ 2A=4{ r }^{ 2 } } $$

$$\Longrightarrow \dfrac { { r }^{ 4 }-4{ r }^{ 2 }+16 }{ 8{ r }^{ 2 } } =\cos { 2A } $$

$$\Longrightarrow -1\le \dfrac { { r }^{ 4 }-4{ r }^{ 2 }+16 }{ 8{ r }^{ 2 } } \le 1$$ ......($$\because \quad -1\le \cos { 2A\le 1 } $$}

$$\Longrightarrow -1\le \dfrac { { r }^{ 4 }-4{ r }^{ 2 }+16 }{ 8{ r }^{ 2 } } \le 1$$
  
Case 1:
$$\dfrac { { r }^{ 4 }-4{ r }^{ 2 }+16 }{ 8{ r }^{ 2 } } \ge 1$$

$$\therefore \quad r\in { R }^{ + }$$        ...{$$\because \quad r>0$$}

Case 2:
$$\dfrac { { r }^{ 4 }-4{ r }^{ 2 }+16 }{ 8{ r }^{ 2 } } \le 1$$

$$\Longrightarrow { r }^{ 4 }-12{ r }^{ 2 }+16\le 0$$

$$\Longrightarrow 6-2\sqrt { 5 } \le { r }^{ 2 }\le 6+2\sqrt { 5 } $$

$$\Longrightarrow { \left( \sqrt { 5 } -1 \right)  }^{ 2 }\le { r }^{ 2 }\le { \left( \sqrt { 5 } +1 \right)  }^{ 2 }$$

$$\therefore \quad \sqrt { 5 } -1\le r\le \sqrt { 5 } +1$$       ....{$$\because \quad r>0$$}

Therefore the difference between maximum and least value is $$\sqrt { 5 } +1-\sqrt { 5 } +1=2$$

Ans: A

Given:-

Given the conditions in the question, the maximum values of $$z_1, z_2,z_3$$ are 2, 4 and 6 respectively.
Thus, the maximum value of $$|z_1 + z_2 + z_3|$$ becomes 12 when all the values are maximum simultaneously.
So, $$|z_1 + z_2 + z_3|$$ is less than 12.

Let $$z=x+iy$$
$$z^{2}$$ is purely imaginary
$$z^{2}=x^{2}-y^{2}+2xyi$$
$$x=\pm y$$
so 4 pair can possible for constant modulus.
$$(x,y),(-x,-y),(x,-y),(-x,y)$$

$$E = \left|z\right| + \left|z - 1\right| + \left|2z - 3\right|
= \left|z\right| + \left|z-1\right| + \left|3-2z\right| \ge \left|z + z - 1+ 3 - 2z\right| = 2$$
$$\therefore, \left|z\right| + \left|z - 1\right| + \left|2z - 3\right| \ge 2$$
Therefore, minimum value is 2.

Ans: B

$$|z_1+z_2|⩽||z_1|+|z_2||$$
$$2 = \left|\left|z - 3\right| - \left|z + 1\right|\right| \le \left|(z-3) + (z+1)\right|$$
$$\Rightarrow \left|z - 1\right|  \ge 1$$

Ans: B 

Given, $$\left|z_1 +  z_2 + z_3\right|$$