Answer :

$[i^{18}+(\dfrac{1}{i})^{25}]^{3}$

$=[i^{4 \times 4+2}+\dfrac{1}{i^{4 \times 6+1}}]^{3}$

$=[(i^{4})^{4} \cdot i^{2}+\dfrac{1}{(i^{4})^{6} \cdot i}]^{3}$

$=[i^{2}+\dfrac{1}{i}]^{3} \quad[i^{4}=1]$

$=[-1+\dfrac{1}{i} \times \dfrac{i}{i}]^{3} \quad[i^{2}=-1]$

$=[-1+\dfrac{i}{i^{2}}]^{3}$

$=[-1-i]^{3}$

$=(-1)^{3}[1+i]^{3}$

$=-[1^{3}+i^{3}+3 \cdot 1 \cdot i(1+i)]$

$=-[1+i^{3}+3 i+3 i^{2}]$

$=-[1-i+3 i-3]$

$=-[-2+2 i]$

$=2-2 i$

Answer :

Let $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$

$\therefore z_1 z_2=(x_1+i y_1)(x_2+i y_2)$

$=x_1(x_2+i y_2)+i y_1(x_2+i y_2)$

$=x_1 x_2+i x_1 y_2+i y_1 x_2+i^{2} y_1 y_2$

$=x_1 x_2+i x_1 y_2+i y_1 x_2-y_1 y_2 \quad[i^{2}=-1]$

$=(x_1 x_2-y_1 y_2)+i(x_1 y_2+y_1 x_2)$

$\Rightarrow Re(z_1 z_2)=x_1 x_2-y_1 y_2$

$\Rightarrow Re(z_1 z_2)=Re z_1 Re z_2-Im z_1 Im z_2$

Hence, proved.

Answer :

$(\dfrac{1}{1-4 i}-\dfrac{2}{1+i})(\dfrac{3-4 i}{5+i})=[\dfrac{(1+i)-2(1-4 i)}{(1-4 i)(1+i)}][\dfrac{3-4 i}{5+i}]$

$=[\dfrac{1+i-2+8 i}{1+i-4 i-4 i^{2}}][\dfrac{3-4 i}{5+i}]=[\dfrac{-1+9 i}{5-3 i}][\dfrac{3-4 i}{5+i}]$

$=[\dfrac{-3+4 i+27 i-36 i^{2}}{25+5 i-15 i-3 i^{2}}]=\dfrac{33+31 i}{28-10 i}=\dfrac{33+31 i}{2(14-5 i)}$

$=\dfrac{(33+31 i)}{2(14-5 i)} \times \dfrac{(14+5 i)}{(14+5 i)} \quad$ [On multiplying numerator and denominator by $(14+5 i)$ ]

$=\dfrac{462+165 i+434 i+155 i^{2}}{2[(14)^{2}-(5 i)^{2}]}=\dfrac{307+599 i}{2(196-25 i^{2})}$

$=\dfrac{307+599 i}{2(221)}=\dfrac{307+599 i}{442}=\dfrac{307}{442}+\dfrac{599 i}{442}$

This is the required standard form.

Answer :

$x-i y=\sqrt{\dfrac{a-i b}{c-i d}}$

$=\sqrt{\dfrac{a-ib}{c-id} \times \dfrac{c+id}{c+id}}[$ On multiplying numerator and deno min ator by $(c+id)]$

$=\sqrt{\dfrac{(ac+bd)+i(ad-bc)}{c^{2}+d^{2}}}$

$\therefore(x-i y)^{2}=\dfrac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}$

$\Rightarrow x^{2}-y^{2}-2 ixy=\dfrac{(ac+bd)+i(ad-bc)}{c^{2}+d^{2}}$

On comparing real and imaginary parts, we obtain

$x^{2}-y^{2}=\dfrac{a c+b d}{c^{2}+d^{2}},-2 x y=\dfrac{a d-b c}{c^{2}+d^{2}}$

$(x^{2}+y^{2})^{2}=(x^{2}-y^{2})^{2}+4 x^{2} y^{2}$

$=(\dfrac{a c+b d}{c^{2}+d^{2}})^{2}+(\dfrac{a d-b c}{c^{2}+d^{2}})^{2} \quad[U sing(1)]$

$=\dfrac{a^{2} c^{2}+b^{2} d^{2}+2 a c b d+a^{2} d^{2}+b^{2} c^{2}-2 a d b c}{(c^{2}+d^{2})^{2}}$

$=\dfrac{a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}}{(c^{2}+d^{2})^{2}}$

$=\dfrac{a^{2}(c^{2}+d^{2})+b^{2}(c^{2}+d^{2})}{(c^{2}+d^{2})^{2}}$

$=\dfrac{(c^{2}+d^{2})(a^{2}+b^{2})}{(c^{2}+d^{2})^{2}}$

$=\dfrac{a^{2}+b^{2}}{c^{2}+d^{2}}$

Hence, proved.

Answer :

$ \begin{aligned} & z_1=2-i, z_2=1+i \\ & \therefore|\dfrac{z_1+z_2+1}{z_1-z_2+1}|=|\dfrac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}| \\ & =|\dfrac{4}{2-2 i}|=|\dfrac{4}{2(1-i)}| \\ & =|\dfrac{2}{1-i} \times \dfrac{1+i}{1+i}|=|\dfrac{2(1+i)}{1^{2}-i^{2}}| \\ & =|\dfrac{2(1+i)}{1+1}| \quad[i^{2}=-1] \\ & =|\dfrac{2(1+i)}{2}| \\ & =|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2} \end{aligned} $

Thus, the value of $|\dfrac{z_1+z_2+1}{z_1-z_2+1}|$ is $\sqrt{2}$.

Answer :

$ \begin{aligned} a+i b & =\dfrac{(x+i)^2}{2 x^2+1} \\ & =\dfrac{x^2+i^2+2 x i}{2 x^2+1} \\ & =\dfrac{x^2-1+i 2 x}{2 x^2+1} \\ & =\dfrac{x^2-1}{2 x^2+1}+i(\dfrac{2 x}{2 x^2+1}) \end{aligned} $

On comparing real and imaginary parts, we obtain

$ \begin{aligned} & a=\dfrac{x^{2}-1}{2 x^{2}+1} \text{ and } b=\dfrac{2 x}{2 x^{2}+1} \\ & \begin{aligned} \therefore a^{2}+b^{2} & =(\dfrac{x^{2}-1}{2 x^{2}+1})^{2}+(\dfrac{2 x}{2 x^{2}+1})^{2} \\ = & \dfrac{x^{4}+1-2 x^{2}+4 x^{2}}{(2 x+1)^{2}} \\ & =\dfrac{x^{4}+1+2 x^{2}}{(2 x^{2}+1)^{2}} \\ & =\dfrac{(x^{2}+1)^{2}}{(2 x^{2}+1)^{2}} \\ \therefore a^{2}+b^{2} & =\dfrac{(x^{2}+1)^{2}}{(2 x^{2}+1)^{2}} \end{aligned} \end{aligned} $

Hence, proved.

Answer :

$z_1=2-i, z_2=-2+i$

(i) $z_1 z_2=(2-i)(-2+i)=-4+2 i+2 i-i^{2}=-4+4 i-(-1)=-3+4 i$

$ \overline{z}_1=2+i $

$\therefore \dfrac{z_1 z_2}{\overline{z}_1}=\dfrac{-3+4 i}{2+i}$

On multiplying numerator and denominator by $( 2 - i)$, we obtain

$ \begin{aligned} \dfrac{z_1 z_2}{\bar{z}_1} & =\dfrac{(-3+4 i)(2-i)}{(2+i)(2-i)}=\dfrac{-6+3 i+8 i-4 i^{2}}{2^{2}+1^{2}}=\dfrac{-6+11 i-4(-1)}{2^{2}+1^{2}} \\ & =\dfrac{-2+11 i}{5}=\dfrac{-2}{5}+\dfrac{11}{5} i \end{aligned} $

On comparing real parts, we obtain

$Re(\dfrac{z_1 z_2}{\bar{z}_1})=\dfrac{-2}{5}$

(ii)

$ \dfrac{1}{z_1 \bar{z}_1}=\dfrac{1}{(2-i)(2+i)}=\dfrac{1}{(2)^{2}+(1)^{2}}=\dfrac{1}{5} $

On comparing imaginary parts, we obtain

$Im(\dfrac{1}{z_1 \bar{z}_1})=0$

Answer :

Let $z=(x-i y)(3+5 i)$

$z=3 x+5 x i-3 y i-5 y i^{2}=3 x+5 x i-3 y i+5 y=(3 x+5 y)+i(5 x-3 y)$

$\therefore \bar{{}z}=(3 x+5 y)-i(5 x-3 y)$

It is given that, $\bar{{}z}=-6-24 i$

$\therefore(3 x+5 y)-i(5 x-3 y)=-6-24 i$

Equating real and imaginary parts, we obtain

$3 x+5 y=-6$ $5 x-3 y=24$

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain

$ \begin{aligned} 9x + 15y &= -18 \\ 25x - 15y &= 120 \\ \hline 34x &= 102 \\ \therefore x &= \dfrac{102}{34} = 3 \end{aligned} $

Putting the value of $x$ in equation (i), we obtain

$ \begin{aligned} & 3(3)+5 y=-6 \\ & \Rightarrow 5 y=-6-9=-15 \\ & \Rightarrow y=-3 \end{aligned} $

Thus, the values of $x$ and $y$ are 3 and - 3 respectively.

Answer :

$ \begin{aligned} \dfrac{1+i}{1-i}-\dfrac{1-i}{1+i} & =\dfrac{(1+i)^{2}-(1-i)^{2}}{(1-i)(1+i)} \\ & =\dfrac{1+i^{2}+2 i-1-i^{2}+2 i}{1^{2}+1^{2}} \\ & =\dfrac{4 i}{2}=2 i \\ \therefore|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}| & =|2 i|=\sqrt{2^{2}}=2 \end{aligned} $

Answer :

$ \begin{aligned} & (x+i y)^{3}=u+i v \\ & \Rightarrow x^{3}+(i y)^{3}+3 \cdot x \cdot i y(x+i y)=u+i v \\ & \Rightarrow x^{3}+i^{3} y^{3}+3 x^{2} y i+3 x y^{2} i^{2}=u+i v \\ & \Rightarrow x^{3}-i y^{3}+3 x^{2} y i-3 x y^{2}=u+i v \\ & \Rightarrow(x^{3}-3 x y^{2})+i(3 x^{2} y-y^{3})=u+i v \end{aligned} $

On equating real and imaginary parts, we obtain

$ \begin{aligned} & u=x^{3}-3 x y^{2}, v=3 x^{2} y-y^{3} \\ & \therefore \dfrac{u}{x}+\dfrac{v}{y}=\dfrac{x^{3}-3 x y^{2}}{x}+\dfrac{3 x^{2} y-y^{3}}{y} \\ & =\dfrac{x(x^{2}-3 y^{2})}{x}+\dfrac{y(3 x^{2}-y^{2})}{y} \\ & =x^{2}-3 y^{2}+3 x^{2}-y^{2} \\ & =4 x^{2}-4 y^{2} \\ & =4(x^{2}-y^{2}) \\ & \therefore \dfrac{u}{x}+\dfrac{v}{y}=4(x^{2}-y^{2}) \end{aligned} $

Hence, proved.

Answer :

Let $\alpha=a+i b $ and $ \beta=x+i y $

It is given that, $|\beta|=1$

$\therefore \sqrt{x^{2}+y^{2}}=1$

$\Rightarrow x^{2}+y^{2}=1$

$ \begin{aligned} &|\dfrac{\beta-\alpha}{1-\bar{{}\alpha} \beta}|=|\dfrac{(x+i y)-(a+i b)}{1-(a-i b)(x+i y)}| \\ &=|\dfrac{(x-a)+i(y-b)}{1-(a x+a i y-i b x+b y)}| \\ &=|\dfrac{(x-a)+i(y-b)}{(1-a x-b y)+i(b x-a y)}| \\ &=\dfrac{|(x-a)+i(y-b)|}{(1-a x-b y)+i(b x-a y) \mid} \quad[|\dfrac{z_1}{z_2}|=\dfrac{|z_1|}{|z_2|}] \\ &=\dfrac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{(1-a x-b y)^{2}+(b x-a y)^{2}}} \\ &=\dfrac{\sqrt{x^{2}+a^{2}-2 a x+y^{2}+b^{2}-2 b y}}{\sqrt{1+a^{2} x^{2}+b^{2} y^{2}-2 a x+2 a b x y-2 b y+b^{2} x^{2}+a^{2} y^{2}-2 a b x y}} \\ &=\dfrac{\sqrt{(x^{2}+y^{2})+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}(x^{2}+y^{2})+b^{2}(y^{2}+x^{2})-2 a x-2 b y}} \\ &=\dfrac{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}} \\ &=1 \\ & \therefore|\dfrac{\beta-\alpha}{1-\bar{{}\alpha} \beta}|=1 \end{aligned} $

Answer :

$ \begin{aligned} & |1-i|^{x}=2^{x} \\ & \Rightarrow(\sqrt{1^{2}+(-1)^{2}})^{x}=2^{x} \\ & \Rightarrow(\sqrt{2})^{x}=2^{x} \\ & \Rightarrow 2^{\dfrac{x}{2}}=2^{x} \\ & \Rightarrow \dfrac{x}{2}=x \\ & \Rightarrow x=2 x \\ & \Rightarrow 2 x-x=0 \\ & \Rightarrow x=0 \end{aligned} $

Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0 .

Answer :

$(a+i b)(c+i d)(e+i f)(g+i h)=A+i B$

$\therefore|(a+i b)(c+i d)(e+i f)(g+i h)|=|A+i B|$

$\Rightarrow|(a+i b)| \times(c+i d)|\times|(e+i f)|\times|(g+i h)|=| A+i B \mid \quad[|z_1 z_2|=|z_1||z_2|]$

$\Rightarrow \sqrt{a^{2}+b^{2}} \times \sqrt{c^{2}+d^{2}} \times \sqrt{e^{2}+f^{2}} \times \sqrt{g^{2}+h^{2}}=\sqrt{A^{2}+B^{2}}$

On squaring both sides, we obtain

$(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2})=A^{2}+B^{2}$

Hence, proved.

Answer :

$ \begin{aligned} & (\dfrac{1+i}{1-i})^{m}=1 \\ & \Rightarrow(\dfrac{1+i}{1-i} \times \dfrac{1+i}{1+i})^{m}=1 \\ & \Rightarrow(\dfrac{(1+i)^{2}}{1^{2}+1^{2}})^{m}=1 \\ & \Rightarrow(\dfrac{1^{2}+i^{2}+2 i}{2})^{m}=1 \\ & \Rightarrow(\dfrac{1-1+2 i}{2})^{m}=1 \\ & \Rightarrow(\dfrac{2 i}{2})^{m}=1 \\ & \Rightarrow i^{m}=1 \end{aligned} $

$\therefore m=4 k$, where $k$ is some integer.

Therefore, the least positive integer is 1.

Thus, the least positive integral value of $m$ is $4(=4 \times 1)$.