Complex Numbers...

The number of solutions of $$log _{\frac{1}{5}}log_{\frac{1}{2}}(\left | z \right |^{2}+4\left | z \right |+3)< 0$$ is/are?

Let $$z$$ be a complex number and $$c$$ be a real number $$\geq $$ 1 such that z + $$c\left | z+1 \right |+i=0 ,$$ then $$c$$ belongs to 

lf $$\displaystyle \log_{\tan 30^{\circ}}\left(\frac{2|Z|^{2}+2|Z|-3}{|z|+1}\right) <-2$$ then

If $$x = 2 + 5i($$where $$1 i = \sqrt{-1})$$ and $$2(\displaystyle \frac{1}{1! 9!}  + \frac{1}{3! 7!}) + \frac{1}{5! 5!} = \frac{2^{a}}{b!}$$ then $$ x^{3}-5x^{2}+33x-10 = $$

If $$b_1b_2=2(c_1+c_2)$$, then at least one of the equations $$x^2+b_1x+c_1=0$$ and $$x^2+b_2x+c_2=0$$ has

If $$x=9^{\frac {1}{3}} 9^{\frac {1}{9}} 9^{\frac {1}{27}} .....\infty, y=4^{\frac {1}{3}} 4^{\frac {-1}{9}} 4^{\frac {1}{27}}....\infty,$$ and $$z=\sum_{r=1}^{\infty} (1+i)^{-r}$$, then $$arg (x+yz)$$ is equal to

The argument of the complex number $$\sin \frac{6\pi }{5}+i\left ( 1+\cos \frac{6\pi }{5} \right )$$ is

Find the value of $$x$$ such that $$\displaystyle \frac{(x + \alpha)^2 - (x + \beta)^2}{ \alpha + \beta} = \frac{sin  2 \theta}{sin^2  \theta}$$. when $$\alpha$$ and $$\beta $$ are the roots of $$t^2 - 2t + 2 = 0$$

The complex numbers $$\sin  x + i  \cos  2x$$  and  $$\cos  x - i  \sin  2x$$ are conjugate to each other, for