Number Theory Test 18

$$z=x+iy = r\cos\theta + ir\sin\theta where \left|z\right|=r$$
Now, $$\left|x\right|+ \left|y\right|=\left| r\cos\theta\right|+\left| r\sin\theta \right|$$
$$=|r|\left \{ \cos\theta|+|\sin\theta| \right \}$$
But $$-\sqrt{2} < \cos s \theta +\ sin\theta < \sqrt2$$
$$\Rightarrow |x|+|y|\leqslant \sqrt{2}|r|$$
$$\leqslant \sqrt{2}|z|$$

Let us assume that there exists a $$z$$ which satisfies the given equation and $$ |z|<1$$
Now,  $${ z }^{ 4 }+z+2=0$$
$$\Rightarrow -2={ z }^{ 4 }+z$$
$$\Rightarrow |-2|=|{ z }^{ 4 }+z|$$
$$ \Rightarrow 2\le |{ z }^{ 4 }|+|z|$$
$$\Rightarrow 2<1+1$$,because$$\quad |z|<1,$$
This is not possible. Hence the equation does not have a root which satisfies $$\quad |z|<1$$
Hence statement 1 is true and Statement 2 is the correct explanation for it.

We have $$(\dfrac{1+ia}{1-ia})^4 = z$$ is a fixed complex number
$$\Rightarrow (\dfrac{1+ia}{1-ia})^4 = cos\theta + isin \theta$$
$$\dfrac{1+ia}{1-ia} = cos(\dfrac{2n\pi+\theta}{4})$$$$+isin\dfrac{2n\pi+\theta}{4}$$

$$\Rightarrow \dfrac{1+ia}{1-ia}=cos\theta + isin\theta$$
Where,$$\phi = \dfrac{2n\phi+\phi}{4}$$
applying componendo dividend,
$$\dfrac{1+ia+1-ia}{1+ia=1+ia}=\dfrac{cos\theta+isin\phi+1}{cos\theta+isin\phi-1}$$
$$\Rightarrow \dfrac{2}{2ia}=\dfrac{cos\theta+isin\phi+1}{cos\theta+isin\phi-1}$$
$$\Rightarrow \dfrac{1}{ia}=\dfrac{2\omega s^2\dfrac{\phi}{2}+2isin\dfrac{\phi}{2}\cos\dfrac{\phi}{2}}{-2sin^2\dfrac{\phi}{2}+2isin\dfrac{\phi}{2}\cos\dfrac{\phi}{2}}$$
$$=co+\dfrac{\theta}{2}\dfrac{(\omega s \dfrac{\theta}{2}+isin\dfrac{\theta}{2})}{(-sin \dfrac{\theta}{2}+isin\dfrac{\theta}{2})}$$
$$\Rightarrow a = -tan \dfrac{\theta}{2}$$
$$= -tan \dfrac{2n\pi + \theta}{8}$$
$$n=0, 1, 2, 3$$
$$i.e a = -tan \dfrac{\theta}{8}, -tan (\dfrac{\pi}{4}+\dfrac{\theta}{8}),$$$$-tan (\dfrac{\pi}{2}+\dfrac{\theta}{8}),$$
$$4 real roots, -tan (\dfrac{3\pi}{2}+\dfrac{\theta}{8}) \dfrac{1}{ia}$$
$$= i cot \dfrac{\theta}{2}$$

$$\dfrac{x+3i}{2+iy}=1-i$$

$$a_{n}=(\sqrt{3+i})(\sqrt{3}-i)^{n-1}$$

$$\downarrow$$            $$\downarrow $$

first       ratio

$$|a_{n}|=|\sqrt{3}+i|(|\sqrt{3-i}|)^{n-1}$$

$$[as |z^{n-1}|=(|z|)^{n-1}]$$

$$=(\sqrt{3+1})\times(\sqrt{3+1})^{n-1}$$

$$=2\times2^{n-1} = 2^n$$

$$\alpha =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =5$$
$$\beta =\sqrt{(5)^{2}+(12)^{2}}=13$$
$$\gamma =\sqrt{1+1^{2}}=\sqrt{2}$$
$$\Rightarrow \gamma < \alpha < \beta$$

Given $$|z-4| < |z-2|$$
all complex numbers
lie farther from $$(2,0)$$
than from $$(4,0)$$ as
$$|z-4|<|z-2|$$
So they lie on the side of $$(z,0)$$ which is
closer to $$(4,0)$$ than $$(2,0)$$
$$z=x+iy \Rightarrow x>3$$
$$\therefore Re (z) >3$$

$$\left|z\right| = z + 1 +2i$$

We have, 
$$\left|w\right| = 1
\Longrightarrow \left|\dfrac{1 - iz}{z - i}\right| = 1$$
or $$\dfrac{\left|1 - iz\right|}{\left|z - i\right|} =1$$
or $$\left|1 - iz\right| = \left|z - i\right|$$       (1)
or $$\left| - i (x + iy) \right| = \left|x + iy - i\right|$$, where $$ z = x + iy$$
or $$\left|1 + y - ix\right| = \left|x + i(y - 1)\right|$$
or $$\sqrt{(1+y)^2 + (-x)^2} = \sqrt{x^2 + (y - 1)^1}$$
or $$(1 + y)^2 + x^2 + x^2 + (y-1)^2$$
or $$y = 0$$
$$\Longrightarrow z = x + i0 = x,$$ which is purely real

Ans: D