Free Objective Test 02 Practice Test - 11th and 12th 

$\frac{1+7i}{(2-i{)}^2}$=$\frac{(1+7i)(3+4i)}{(3-4i)(3+4i)}$=$\frac{-25+25i}{25}$= -1+i

Let z=x+iy = -1+i

$\therefore$ rcos$\theta$ =-1 And rsin$\theta$ =1 $\therefore$ $\theta$=$\frac{3\pi }{4}$ and r=$\sqrt{2}$

Thus $\frac{1+7i}{(2-i{)}^2}$=$\sqrt{2}$ $[cos\frac{3\pi }{4}+isin\frac{3\pi }{4}]$

Alter: $\left|\frac{1+7i}{(2-i{)}^2}\right|$=$\left|\frac{1+7i}{3-4i}\right|$=$\sqrt{2}$

And arg $\left(\frac{1+7i}{(2-i{)}^2}\right)$=${\tan }^{-1}7-{\tan }^{-1}\left(\frac{-4}{3}\right)$

                                                                    =${\tan }^{-1}7+{\tan }^{-1}\left(\frac{4}{3}\right)$=$\frac{3\pi }{4}$

$\therefore$ $\frac{1+7i}{(2-i{)}^2}$=$\sqrt{2}$ $[cos\frac{3\pi }{4}+isin\frac{3\pi }{4}]$

z = 3+5i , $\overline{z}$ = 3-5i

⇒ $z^3=(3+5i{)}^3=3^3+(5i{)}^3+\mathrm{3.3.5}i(3+5i)$

              = -198+10i

Hence, $z^3+\overline{z}$ + 198 = 10i-198+3-5i+198 =  3+5i . 

${(\frac{i-1}{i+1}\times \frac{i-1}{i-1})}^n={\left(\frac{-2i}{-2}\right)}^n=i^n$  

Hence, to make the real number the least positive integar is 2.

We have a $=\sqrt{2i}=\sqrt{2}{(cos\frac{\pi }{2}+isin\frac{\pi }{2})}^{\frac{1}{2}}=\sqrt{2}{\left(e^{i^{\frac{\pi }{2}}}\right)}^{\frac{1}{2}}=\sqrt{2}{e}^{i^{\frac{\pi }{4}}}=\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}i)=1+i$
Trick:Check with options.

Let z = $i^{[1+3+5+......+(2n+1\left)\right]}$

Clearly series is A.P with common difference = 2 

∵  $T_n$ = 2n-1 And  $T_{n+1}$ = 2n+1

So, number of terms in A.P. = n+1

Now, $s_{n+1}=\frac{n+1}{2}$ [2.1+(n+1-1)2] 

$\Rightarrow$ $S_{n+1}=\frac{n+1}{2}[2+2n]=(n+1{)}^2$ i.e., $i^{(n+1{)}^2}$

Now put n = 1,2,3,4,5,........

n = 1, z = $i^4$ = 1, n = 2, z = $i^5$  = -1 , 

n = 3, z = $i^8$ = 1, n = 4, z = $i^{10}$  = -1 , 

n = 5, z = $i^{12}$ = 1, .........

     Let z = x + iy, $\overline{z}$ = x -iy  and  $z^{-1}=\frac{1}{x+iy}$

⇒ $\overline{z^{-1}}=\frac{x+iy}{x^2+y^2}$ ; ∴ $\overline{z^{-1}}\overline{z}=\frac{x+iy}{x^2+y^2}(x-iy)$  = 1 

sinx+icos2x and cosx-isin2xare conjugate to each other if sinx=cosx and cos2x=sin2x

Or     tan x = 1   ⇒ x   = $\frac{\pi }{4}$,$\frac{5\pi }{4}$,$\frac{9\pi }{4}$, ..........            ...............(i)

And  tan 2x = 1 ⇒ 2x = $\frac{\pi }{4}$,$\frac{5\pi }{4}$,$\frac{9\pi }{4}$, ..........

or                              x  =  $\frac{\pi }{8}$,$\frac{5\pi }{8}$,$\frac{9\pi }{8}$, ..........           ...............(ii)

There exists no value of x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate.

$\frac{b}{c}$=$\frac{cos\beta +isin\beta }{cos\gamma +isin\gamma }\times \frac{cos\gamma -isin\gamma }{cos\gamma -isin\gamma }$

$\Rightarrow \frac{b}{c}=cos(\beta -\gamma )+isin(\beta -\gamma ).....\left(i\right)$

Similarly, $\frac{c}{a}=cos(\gamma -\alpha )+isin(\gamma -\alpha ).......\left(ii\right)$

And $\frac{a}{b}=cos(\alpha -\beta )+isin(\alpha -\beta ).......\left(ii\right)$

Equating real and imaginary parts,

$cos(\beta -\gamma )+cos(\gamma -\alpha )+cos(\alpha -\beta )=1$

cos$\alpha$+cos$\beta$+cos$\gamma$=0 ......(i)

sin$\alpha$+sin$\beta$+sin$\gamma$=0 .......(ii)

 Let a=cos$\alpha$+isin$\alpha$, b=cos$\beta$+isin$\beta$

      c=cos$\gamma$+isin$\gamma$

$\Rightarrow$ a+b+c=0 .......(iii)

 $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$

=${[cos\alpha +isin\alpha ]}^{-1}$+${[cos\beta +isin\beta ]}^{-1}$+${[cos\gamma +isin\gamma ]}^{-1}$

=cos$\alpha$-isin$\alpha$+cos$\beta$-isin$\beta$+cos$\gamma$-isin$\gamma$

$\Rightarrow$ ab+bc+ca=0 .....(iv) [by (i) and (ii)]

Squaring both sides of equations (iii),

We get $a^2$+$b^2$+$c^2$+2ab+2bc+2ca=0

or $a^2$+$b^2$+$c^2$ [by (iv)]

 ${[cos\alpha +isin\alpha ]}^2$+${[cos\beta +isin\beta ]}^2$+${[cos\gamma +isin\gamma ]}^2$=0

$\Rightarrow$ (cos2$\alpha$+cos2$\beta$+cos2$\gamma$) + i(sin2$\alpha$+sin2$\beta$+sin2$\gamma$)

Equating the real part to zero, we have

$\Rightarrow cos2\alpha +cos2\beta +cos2\gamma =0$

cos$\alpha$+cos$\beta$+cos$\gamma$ = 0............(i)

 sin$\alpha$+sin$\beta$+sin$\gamma$  =0...........(ii)

Let a=cos$\alpha$+isin$\alpha$, b=cos$\beta$+isin$\beta$

     c=cos$\gamma$+isin$\gamma$

$\Rightarrow$ a+b+c=0 [By (i) and (ii)]....(iii)

$\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$

=cos$\alpha$-isin$\alpha$+cos$\beta$-isin$\beta$+cos$\gamma$-isin$\gamma$

$\Rightarrow$ ab+bc+ca=0 ......(iv) [by (i) and (ii)]

Squaring both sides of equation (iii),

we get $a^2$+$b^2$+$c^2$+2ab+2bc+2ca=0

or $a^2$+$b^2$+$c^2$=0     [by (iv)]

 $[cos\alpha +isin\alpha {]}^2$ + $[cos\beta +isin\beta {]}^2$ + $[cos\gamma +isin\gamma {]}^2$ = 0

$\Rightarrow$ (cos2$\alpha$+cos2$\beta$+cos2$\gamma$) + i(sin2$\alpha$+sin2$\beta$+sin2$\gamma$)=0

Separation of real and imaginary part,

$\Rightarrow$ cos2$\alpha$+cos2$\beta$+cos2$\gamma$=0 .........(v)

And sin2$\alpha$+sin2$\beta$+sin2$\gamma$ =0 .........(iv)

$\Rightarrow$ 1-2$si{n}^2$$\alpha$+1-2$si{n}^2$$\beta$+1-2$si{n}^2$$\gamma$=0 [By eq.(v)]

Therefore, $si{n}^2$$\alpha$+$si{n}^2$$\beta$+$si{n}^2$$\gamma$= $\frac{3}{2}$

let z=r (cos$\theta$+isin$\theta$)

Then $|z+\frac{1}{z}|$=a $\Rightarrow$ ${|z+\frac{1}{z}|}^2$=$a^2$

$\Rightarrow$ $r^2$+$\frac{1}{r^2}$+2cos$\theta$ = $a^2$     $\cdots$ $\cdots$(i)

Differentiating w.r.t $\theta$ we get

2r$\frac{dr}{d\theta }$-$\frac{2}{r^3}$$\frac{dr}{d\theta }$-4sin2$\theta$

Putting $\frac{dr}{d\theta }$=0, we get $\theta$=0,$\frac{\pi }{2}$

r is maximum for $\theta$ = 0, $\frac{\pi }{2}$, therefore from (i)

$r^2+\frac{1}{r^2}-2=a^2\Rightarrow r-\frac{1}{r}$=$a\Rightarrow r$=$\frac{a+\sqrt[2]{2a^2+4}}{2}$

 $\sqrt{3}+i$=(a+ib)(c+id)

$\therefore ac-bd$=$\sqrt{3}$ and ad+bc=1

Now  ${\tan }^{-1}\left(\frac{b}{a}\right)+{\tan }^{-1}\left(\frac{d}{c}\right)$

=  ${\tan }^{-1}\left(\frac{\frac{a}{b}+\frac{d}{c}}{1-\frac{b}{a}.\frac{d}{c}}\right)$= ${\tan }^{-1}\left(\frac{bc+ad}{ac-bd}\right)$= ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

=n$\pi$+$\frac{\pi }{6}$, n$ϵ$I

Since1-i = $\sqrt{2}$ $[cos\frac{\pi }{4}-isin\frac{\pi }{4}]$, |1-i|

$\therefore$ $|1-i{|}^x$=$2^x$ $\Rightarrow$ ${\left(\sqrt{2}\right)}^x$=$2^x$ $\Rightarrow$ $2^{x/2}$ =$2^x$

$\Rightarrow$ $\frac{x}{2}$=x then x=0.

Therefore, the number of non-zero integral solutions is nil or Zero.

      $\frac{c+i}{c-i}$ = a+ib     .........(i)

∴   $\frac{c-i}{c+i}$ = a-ib     .........(ii)

Multiplying (i) and (ii), we get 

$\frac{c^2+1}{c^2+1}=a^2+b^2\Rightarrow a^2+b^2=1$.

Let $\sqrt{3-4i}$=x+iy $\Rightarrow$ 3-4i= $x^2$-$y^2$+2ixy

$\Rightarrow$ $x^2$-$y^2$=3, 2xy=-4   ..........(i)

$\Rightarrow$ ${(x^2+y^2)}^2$=${(x^2-y^2)}^2$+4$x^2$$y^2$=$(3{)}^2$+$(-4{)}^2$=25

$\Rightarrow$ ${(x^2+y^2)}^2$=5           ............(ii)

From equation (i) and (ii) $\Rightarrow$ $x^2$=4 $\Rightarrow$x=$\pm$2,

$y^2$=1 $\Rightarrow$y=$\pm$1.Hence the square root of (3-4i)

is $\pm$(2-i).