Chapter – 5(Miscellaneous Exercise)

Complex Numbers and Quadratic Equations

Chapter 5 Miscellaneous ncert math solution class 11

Question 1: Evaluate: $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}$

Solution : Given $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}$

$=\left(i^{4}\right)^{4} \cdot i^{2}+\left\{\frac{1}{\left(i^{4}\right)^{6} . i}\right\}^{3}$

$=\left[1 \cdot i^{2}+\left\{\frac{1}{1 . i}\right\}\right]^{3} \quad\left[\because i^{4}=1\right]$

$=\left[-1+\frac{1}{i} \times \frac{i}{i}\right]^{3} \quad\left[\because i^{2}=-1\right]$

$=\left[-1+\frac{i}{i^{2}}\right]^{3}\quad\left[\because i^{2}=-1\right]$

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

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

$=-1-i^{3}-3 i-3 i^{2}$

$=-1+i-3 i+3 \quad\left[\because i^{2}=-1\right]$

$=2-2 i$

Question 2: For any two complex numbers $z_{1}$ and $z_{2}$ , prove that:
$\operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2} \text {. }$

Solution: Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$

Therefore, $z_{1} z_{2}=\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)$

$=x_{1} x_{2}+i x_{1} y_{2}+i x_{2} y_{1}+i^{2} y_{1} y_{2}$

$=x_{1} x_{2}+i x_{1} y_{2}+i x_{2} y_{1}-y_{1} y_{2} \quad\left[\because i^{2}=-1\right]$

$=\left(x_{1} x_{2}-y_{1} y_{2}\right)+i\left(x_{1} y_{2}+x_{2} y_{1}\right)$

Now $\operatorname{Re}\left(z_{1} z_{2}\right)=x_{1} x_{2}-y_{1} y_{2}$

$=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}$

Question 3: Reduce $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$ to the standard form.

Solution :Given that $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$

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

$=\left[\frac{1+i-2+8 i}{1+i-4 i-4 i^{2}}\right]\left(\frac{3-4 i}{5+i}\right)$

$=\left[\frac{-1+9 i}{1-3 i+4}\right]\left(\frac{3-4 i}{5+i}\right) \quad\left[\because i^{2}=-1\right]$

$=\left[\frac{-1+9 i}{5-3 i}\right]\left(\frac{3-4 i}{5+i}\right)$

$=\frac{-3+4 i+27 i-36 i^{2}}{25+5 i-15 i-3 i^{2}}$

$=\frac{-3+31 i+36}{25-10 i+3} \quad\left[\because i^{2}=-1\right]$

$=\frac{33+31 i}{28-10 i}$

$=\frac{33+31 i}{2(14-5 i)}$

$=\frac{33+31 i}{2(14-5 i)} \times \frac{(14+5 i)}{(14+5 i)}$

$=\frac{462+165 i+434 i+155 i^{2}}{2\left[14^{2}-(5 i)^{2}\right]}$

$=\frac{307+599 i}{2\left(196-25 i^{2}\right)}$

$=\frac{307+599 i}{2(196+25)}$

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

$=\frac{307+599 i}{442}$

$=\frac{307}{442}+\frac{599}{442} i$

Question 4: If $x-i y=\sqrt{\frac{a-i b}{c-i d}}$ prove that $x^{2}+y^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$

Solution : Given that $x-i y=\sqrt{\frac{a-i b}{c-i d}}$

Squaring both side

$(x-iy)^2=\frac{a-ib}{c-id}--(i)$

Taking conjugate of eq(i) using formula $\overline{(\frac{z_1}{z_2})}=\frac{\overline{z_1}}{\overline{z_2}}$

$(x+iy)^2 =\frac{a+ib}{c+id}--(ii)$

Multiply (i) and (ii)

$(x-iy)^2.(x+iy)^2=\frac{a-ib}{c-id}.\frac{a+ib}{c+id}$

$\Rightarrow [(x-iy)(x+iy)]^2=\dfrac{(a)^2-(ib)^2}{(c)^2-(id)^2}$

$\Rightarrow [x^2-i^2y^2]=\dfrac{a^2-i^2b^2}{c^2-i^2d^2}$

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

Question 5: Convert into polar form:

(i) $\frac{1+7 i}{(2-i)^{2}}$

(ii) $\frac{1+3 i}{1-2 i}$

Solution : (i) $\frac{1+7 i}{(2-i)^{2}}$

$=\frac{1+7 i}{4+i^{2}-4 i}$

$=\frac{1+7 i}{4-1-4 i} \quad\left[\because i^{2}=-1\right]$

$=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}$

$=\frac{3+4 i+21 i+28 i^{2}}{3^{2}-4^{2} i^{2}}$

$=\frac{3+25 i-28}{9+16} \quad\left[\because i^{2}=-1\right]$

$=\frac{-25+25 i}{25}=-1+i$

Let $Z=-1+i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=-1--(i)$

$r \sin \theta=1--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2 \quad \Rightarrow r^{2}=2$

$\Rightarrow r=\sqrt{2} \quad[\because r=|Z|>0]$

Therefore, modulus $=\sqrt{2}$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{1}{-1} \quad \Rightarrow \tan \theta=-1$

From the equations (1), (2) and (3), it is clear that $\cos \theta$ and $\tan \theta$ are negative but $\sin \theta$ is positive. So, $\theta$ lies in II quadrant. Therefore,

$\operatorname{Argument} \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$

Therefore, the polar form of $Z=-1+i$ is given by

$\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$

(ii) $\frac{1+3 i}{1-2 i}=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}$

$=\frac{1+2 i+3 i+6 i^{2}}{1^{2}-2^{2} i^{2}}$

$=\frac{1+5 i-6}{1+4} \quad\left[\because i^{2}=-1\right]$

$=\frac{-5+5 i}{5}=-1+i$

Let $Z=-1+i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=---(i)$

$r \sin \theta=1--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2$

$\Rightarrow r^{2}=2 \quad \Rightarrow r=\sqrt{2}, \quad|Z|>0$

Therefore, modulus $=\sqrt{2}$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{1}{-1} \quad \Rightarrow \tan \theta=-1$

From the equations (1), (2) and (3), it is clear that $\cos \theta$ and $\tan \theta$ are negative but $\sin \theta$ is positive. So, $\theta$ lies in Il quadrant. Therefore,

$\text { Argument } \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$

Therefore, the polar form of $Z=-1+i$ is given by

$\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$

Solve each of the equation in Exercises 6 to 9.

Question 6: $3 x^{2}-4 x+\frac{20}{3}=0$

Solution: Given that $3 x^{2}-4 x+\frac{20}{3}=0$

The given quadratic equation is $3 x^{2}-4 x+\frac{20}{3}=0$

$\Rightarrow 9 x^{2}-12 x+20=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ , we obtain $a=9, b=-12$ , and $c=20$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c=(-12)^{2}-4 \times 9 \times 20$

$=144-720=-576$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-12) \pm \sqrt{-576}}{2 \times 9}$

$=\frac{12 \pm 24 \cdot \sqrt{-1}}{18}$

$=\frac{2 \pm 4 i}{3} \quad[\because \sqrt{-1}=i]$

Question 7: $x^{2}-2 x+\frac{3}{2}=0$ :

Solution : Given that $x^{2}-2 x+\frac{3}{2}=0$

The given quadratic equation is $x^{2}-2 x+\frac{3}{2}=0$

$\Rightarrow 2 x^{2}-4 x+3=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ , we obtain $a=2, b=-4$ , and $c=3$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=(-4)^{2}-4 \times 2 \times 3$

$=16-24=-8$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-4) \pm \sqrt{-8}}{2 \times 2}$

$=\frac{4 \pm 2 \sqrt{2} \cdot \sqrt{-1}}{4}$

$=\frac{2 \pm \sqrt{2} i}{2} \quad[\because \sqrt{-1}=i]$

Question 8: $27 x^{2}-10 x+1=0$

Solution : Given that $27 x^{2}-10 x+1=0$

The given quadratic equation is $27 x^{2}-10 x+1=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ , we obtain $a=27, b=-10$ , and $c=1$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c=(-10)^{2}-4 \times 27 \times 1$

$=100-108=-8$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-10) \pm \sqrt{-8}}{2 \times 27}$

$=\frac{10 \pm 2 \sqrt{2} \cdot \sqrt{-1}}{54}$

$=\frac{5 \pm \sqrt{2} i}{27} \quad[\because \sqrt{-1}=i]$

Question 9: $21 x^{2}-28 x+10=0$

Solution : The given quadratic equation is $21 x^{2}-28 x+10=0$ .

On comparing the given equation with $a x^{2}+b x+c=0$ , we obtain $a=21, b=-28$ , and $c=10$

Therefore, the discriminant of the given equation is given by

$D=b^{2}-4 a c$

$=(-28)^{2}-4 \times 21 \times 10$

$=784-840=-56$

Therefore, the required solutions are

$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-28) \pm \sqrt{-56}}{2 \times 21}$

$=\frac{28 \pm 2 \sqrt{14} \cdot \sqrt{-1}}{42}$

$=\frac{14 \pm \sqrt{14} i}{21} \quad[\because \sqrt{-1}=i]$

Question 10: If $z_{1}=2-i, z_{2}=1+i$ , find $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|$ .

Solution : Since, $\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}$

$=\frac{2-i+1+i+1}{2-i-1-i+1}$

$=\frac{4}{2-2 i}$

$=\frac{2}{1-i} \times \frac{1+i}{1+i}$

$=\frac{2(1+i)}{1-i^{2}}$

$=\frac{2(1+i)}{2}=1+i$

Therefore, $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=|1+i|$

$=\sqrt{1^{2}+1^{2}}=\sqrt{2}$

Question 11: If $a+i b=\frac{(x+i)^{2}}{2 x^{2}+1}$ , prove that $a^{2}+b^{2}=\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}$

Solution: Here, $a+i b=\frac{(x+i)^{2}}{2 x^{2}+1}$

$\Rightarrow a+i b=\frac{x^{2}+i^{2}+2 x i}{2 x^{2}+1}$

$=\frac{x^{2}-1+i 2 x}{2 x^{2}+1}$

$=\frac{x^{2}-1}{2 x^{2}+1}+\frac{2 x}{2 x^{2}+1} i$

$a=\frac{x^{2}-1}{2 x^{2}+1}$ and $b=\frac{2 x}{2 x^{2}+1}$

Now, $a^{2}+b^{2}=\left(\frac{x^{2}-1}{2 x^{2}+1}\right)^{2}+\left(\frac{2 x}{2 x^{2}+1}\right)^{2}$

$=\frac{x^{4}+1-2 x^{2}}{\left(2 x^{2}+1\right)^{2}}+\frac{4 x^{2}}{\left(2 x^{2}+1\right)^{2}}$

$=\frac{x^{4}+1-2 x^{2}+4 x^{2}}{\left(2 x^{2}+1\right)^{2}}$

$=\frac{x^{4}+1+2 x^{2}}{\left(2 x^{2}+1\right)^{2}}$

$a^{2}+b^{2}=\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}$

Question 12: Let $z_{1}=2-i, z_{2}=-2+i$ . Find

(i) $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\overline{z_{1}}}\right)$

(ii) $\operatorname{Im}\left(\frac{1}{z_{1} \overline{z_{1}}}\right)$

Solution : (i) $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\overline{z_{1}}}\right)$

$\frac{z_{1} z_{2}}{\overline{z_{1}}}=\frac{(2-i)(-2+i)}{(2+i)}$

$=\frac{-4+2 i+2 i-i^{2}}{2+i}$

$=\frac{-4+4 i+1}{2+i} \quad\left[\because i^{2}=-1\right]$

$=\frac{-3+4 i}{2+i} \times \frac{2-i}{2-i}$

$=\frac{-6+3 i+8 i-4 i^{2}}{2^{2}-i^{2}}$

$=\frac{-6+11 i+4}{4+1} \quad\left[\because i^{2}=-1\right]$

$=\frac{-2+11 i}{5}=\frac{-2}{5}+\frac{11}{5} i$

Therfore, $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\overline{z_{1}}}\right)$

$=\operatorname{Re}\left(\frac{-2}{5}+\frac{11}{5} i\right)=-\frac{2}{5}$

(ii) $\operatorname{Im}\left(\frac{1}{z_{1} \overline{z_{1}}}\right)$

$\frac{1}{z_{1} \overline{z_{1}}}=\frac{1}{(2-i)(2+i)}$

$=\frac{1}{2^{2}-i^{2}}=\frac{1}{4+1}$

$=\frac{1}{5} \quad\left[\because i^{2}=-1\right]$

Therefore, $\operatorname{Im}\left(\frac{1}{z_{1} \overline{z_{1}}}\right)$

$=\operatorname{Im}\left(\frac{1}{5}+0 i\right)=0$

Question 13: Find the modulus and argument of the complex number $\frac{1+2 i}{1-3 i}$ .

Solution : $\frac{1+2 i}{1-3 i}=\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}$

$=\frac{1+3 i+2 i+6 i^{2}}{1^{2}-3^{2} i^{2}}$

$=\frac{1+5 i-6}{1+9} \quad\left[\because i^{2}=-1\right]$

$\Rightarrow \frac{1+2 i}{1-3 i}$

$=\frac{-5+5 i}{10}$

$=-\frac{1}{2}+\frac{1}{2} i$

Let $Z=-\frac{1}{2}+\frac{1}{2} i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=-\frac{1}{2}--(i)$

$r \sin \theta=\frac{1}{2}--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=\frac{1}{4}+\frac{1}{4}$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{2}{4} \quad \Rightarrow r^{2}=\frac{1}{2}$

$\Rightarrow r=\frac{1}{\sqrt{2}} \quad[\because r=|Z|>0]$

Therefore, modulus $=\frac{1}{\sqrt{2}}$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{\frac{1}{2}}{-\frac{1}{2}} \quad \Rightarrow \tan \theta=-1$

From the equations (1), (2) and (3), it is clear that $\cos \theta$ and $\tan \theta$ are negative but $\sin \theta$ is positive. So, $\theta$ lies in II quadrant. Therefore,

Argument $= \pi - \frac{\pi}{4}$

$=\frac{3\pi}{4}$

Question 14: Find the real numbers $x$ and $y$ if $(x-i y)(3+5 i)$ is the conjugate of $-6-24 i$ .

Solution : Given that: $(x-i y)(3+5 i)$ is the conjugate of $-6-24 i$

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

$\Rightarrow 3 x+x 5 i-3 y i-5 y i^{2}=-6+24 i$

$\Rightarrow 3 x+(5 x-3 y) i+5 y=-6+24 i \quad\left[\because i^{2}=-1\right]$

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

Comparing the real and imaginary parts, we have

$3 x+5 y=-6--(i)$ and $5 x-3 y=24--(ii)$

Form the equation (1), we have $y=\frac{-6-3 x}{5}$

Putting the value of $y$ in equation (2), we get

$5 x-3\left(\frac{-6-3 x}{5}\right)=24 \quad \Rightarrow 25 x+18+9 x=120$

$\Rightarrow 34 x=102 \quad \Rightarrow x=\frac{102}{34}=3$

Putting the value of $x$ in equation (1), we have

$3 \times 3+5 y=-6$

$\Rightarrow 5 y=-6-9=-15$

$\Rightarrow y=-\frac{15}{5}=-3$

Question 15: Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$

Solution : Given that $\frac{1+i}{1-i}-\frac{1-i}{1+i}$

$=\frac{(1+i)(1+i)-(1-i)(1-i)}{(1-i)(1+i)}$

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

$=\frac{4 i}{1+1}=2 i \quad\left[\because i^{2}=-1\right]$

Therefore, $\left|\frac{1+i}{1-i}-\frac{1-i}{1+i}\right|$

$=|2 i|=|0+2 i|$

$=\sqrt{0^{2}+2^{2}}$

$=\sqrt{4}=2$

Question 16: If $(x+i y)^{3}=u+i v$ , then show that $\frac{u}{x}+\frac{v}{y}=4\left(x^{2}-y^{2}\right)$ .

Solution : Given that: $(x+i y)^{3}=u+i v$

$\Rightarrow x^{3}+(i y)^{3}+3 x^{2}(i y)+3 x(i y)^{2}=u+i v$

$\Rightarrow x^{3}-i y^{3}+i 3 x^{2} y-3 x y^{2}=u+i v \quad\left[\because i^{2}=-1\right]$

$\Rightarrow\left(x^{3}-3 x y^{2}\right)+i\left(3 x^{2} y-y^{3}\right)=u+i v$

Comparing the real and imaginary parts, we have

$u=x^{3}-3 x y^{2} \quad$ and $\quad v=3 x^{2} y-y^{3}$

$\Rightarrow \frac{u}{x}=x^{2}-3 y^{2} \quad$ and $\quad \frac{v}{y}=3 x^{2}-y^{2}$

$\Rightarrow \frac{u}{x}+\frac{v}{y}=x^{2}-3 y^{2}+3 x^{2}-y^{2}$

$=4 x^{2}-4 y^{2}=4\left(x^{2}-y^{2}\right)$

Question 17: If $\alpha$ and $\beta$ are different complex number with $|\beta|=1$ , then find $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ .

Solution : We know that $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}$

$=\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right) \overline{\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)}\quad \left[\because z \bar{z}=|z|^{2}\right]$

$=\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}}\right)$

$=\frac{\beta \bar{\beta}-\bar{\alpha} \beta-\alpha \bar{\beta}+\alpha \bar{\alpha}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+\alpha \bar{\alpha} \beta \bar{\beta}}$

$=\frac{|\beta|^{2}-\bar{\alpha} \beta-\alpha \bar{\beta}+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2}|\beta|^{2}} \quad\left[\because z \bar{z}=|z|^{2}\right]$

$=\frac{1-\bar{\alpha} \beta-\alpha \bar{\beta}+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2}} \quad[\because|\beta|=1]$

$=1$

$\Rightarrow\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}=1$

$\Rightarrow\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=1$

Question 18: Find the number of non-zero integral solutions of the equation $|1-i|^{x}=2^{x}$ .

Solution : Given that: $|1-i|^{x}=2^{x}$

$\Rightarrow\left(\sqrt{1^{2}+1^{2}}\right)^{x}=2^{x}$

$\Rightarrow(\sqrt{2})^{x}=2^{x}$

$\Rightarrow 2^{\frac{x}{2}}=2^{x}$

$\Rightarrow \frac{x}{2}=x$

$\Rightarrow x=2 x$

$\Rightarrow x=0$

Thus, 0 is the only integral solution of the given equation.

Therefore, the number of nonzero integral solutions of the given equation is 0 .

Question 19: If $(a+i b)(c+i d)(e+i f)(g+i h)=A+i B$ , then show that $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=A^{2}+B^{2}$ .