TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

$\sum_{r=1}^{16} \left( \sin \frac{2r\pi}{17}+i\cos \frac{2r \pi}{17}\right)=$

Answer:

Explanation:

We have ,

$\sum_{r=1}^{16} \left( \sin \frac{2r\pi}{17}+i\cos \frac{2r \pi}{17}\right)=i\sum_{r=1}^{16}\left( \cos \frac{2r\pi}{17}-i\sin\frac{2r \pi}{17}\right)$

=i$\sum_{r=1}^{16}  e^{-\frac{i2r\pi}{17}}$

$=i\sum_{r=1}^{16}  K^{r}$                           [where,  $K=e^{-\frac{2i \pi}{17}}$]

=$i\left[k\left(\frac{1-k^{16}}{1-k}\right)\right]= i\frac{(k-k^{17})}{(1-k)}$

  =$\frac{i(k-1)}{(1-k)}$

     $[\because k^{17}=e^{-2i\pi}=\cos (2\pi)-i\sin(2\pi)=1]$

  =-i

If P is a complex number whose modulus is one , then the equation $\left(\frac{1+iz}{1-iz}\right)^{4}$  =P has 

Answer:

Explanation:

We have

 $\left(\frac{1+iz}{1-iz}\right)^{4}$  =P

$\Rightarrow$      $\left(\frac{i-z}{i+z}\right)^{4}  =P$

$\Rightarrow$    $\left(\frac{z-i}{z+i}\right)^{4}  =P$

$\Rightarrow$      $|\frac{z-i}{z+i}|^{4}  =|P|$

$\Rightarrow$   $|\frac{z-i}{z+i}|^{4}  =1$

$\Rightarrow$     $|z-i|^{4}= |z+i|^{4}$

$\Rightarrow$     |z-i|=|z+i|

$\therefore$ z lies on perpendicular bisector of i and -i

$\therefore$   z lies on y-axis

 $\therefore$   z has all complex roots 

If z= x+iy represents a point P  in the  argand plane, then the area of the region, represented by the inequality

 2<|z-(1+i)|<3 is 

Answer:

Explanation:

 We have,

     z=x+iy

and 2 <|z-(1+i)|<3

2<|(X-1)-(Y-1)i| <3

 4<(x-1)²+(y-1)² <9

 Area of region= Area of largest circle 

 Area of smallest circle 

$9\pi-4\pi$=$5\pi$

If   $z=\sqrt{2}\sqrt{1+\sqrt{3i}}$  repesents  a point P in the argand  plane and P lies in the third  quadrant , then the polar form of z is 

Answer:

Explanation:

We have

  $z= \sqrt{2}\sqrt{1+\sqrt{3i}}\Rightarrow z=\sqrt{2+2\sqrt{3i}}$

  $z=\sqrt{(\sqrt{3}+i)^{2}}\Rightarrow z=\pm (\sqrt{3}+i)$

 z lie in third  quadrant

$\therefore$    z=$-\sqrt{3}-i$

  $\Rightarrow |z|=\sqrt{(-\sqrt{3})^{2}+(-1)^{2}}=\sqrt{3+1}=\sqrt{4}=2$ and 

$\tan\theta=|\frac{-1}{-\sqrt{3}}|=\frac{1}{\sqrt{3}}\Rightarrow\theta=\frac{\pi}{6}$

 Since   $\theta$ lie in 3rd quadrant

$\therefore$   $arg(z)=-\left(\pi-\frac{\pi}{6}\right)= -\frac{5 \pi}{6}$

Hence,   $z=2\left( \cos \left( -\frac{5 \pi}{6}\right) +i \sin \left(-\frac{5 \pi}{6}\right)\right)$

The value of $\theta$  for which the following  system of equations has a non-trivial solution is  $(4 \sin \theta)x-3y-z=0 , x-(6 \cos 2\theta )y+z=0$, 3x-12y+4z=0

Answer:

Explanation:

We have,

$(4 \sin \theta )x-3y+z=0$

$ x-(6 \cos 2 \theta)y+z=0$

 3x-12y+4z=0

 system of equation has a non trival solution

$\therefore$    $\begin{bmatrix}4 \sin \theta &-3&1 \\1 & -6 \cos 2\theta&1\\3&-12&4 \end{bmatrix}=0$

 $R_{1}\rightarrow 4R_{1}-R_{3}$

$\begin{bmatrix}16 \sin \theta-3 &0&0 \\3 & -6 \cos 2\theta&1\\3&-12&4 \end{bmatrix}=0$

  =   $(16 \sin \theta -3)(-24 \cos 2 \theta+12)=0$

  =$(16 \sin \theta-3)(1-2 \cos \theta)=0$

 $\Rightarrow$       $16 \sin \theta  =1  or 2 \cos  \theta$=1

   $\theta =  \sin^{1}\frac{3}{16}$   or $\theta$ = $\frac{\pi}{6}$

 $\therefore$    $\sin^{-1} (\frac{3}{16})$

• Mathematics - General questions