Ncert Solutions Class 11 Maths Complex Numbers Miscellaneous

\( \left[i^{28}+\left(\frac{1}{i}\right)^{25}\right]^{3} \)
\(=\left[i^{4+4+2}+\frac{1}{(i)^{4+2+1}}\right]^{3}\)
\(=\left[\left(i^{4}\right)^{6} \cdot i^{2}+\frac{1}{\left(i^{4}\right)^{6} \cdot i}\right]^{3}\)
\(=\left[i^{2}+\frac{1}{i}\right]^{3}\left[i^{4}=1\right]\)
\(=\left[-1+\frac{1}{i} \times \frac{i}{i}\right]^{3}\left[i^{2}=-1\right]\)
\(=\left[-1+\frac{i}{i^{2}}\right]^{3}\)
\(=[-1-i]^{3}\)
\(=(-1)^{3}[1+i]^{3}\)
\(=-\left[1^{3}+i^{3}+3 \cdot 1 \cdot i(1+i)\right]\)
\(=-\left[1+i^{3}+3 i+3 i^{2}\right]\)
\(=-[-2+2 \mathrm{i}]\)
\(=2-2 \mathrm{i}\)

It is given in the question that,
\( z_{1} \) and \( z_{2} \) are two complex numbers and we have to prove that:
\( \operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Rez}_{1} \operatorname{Rez}_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2} \)
For this, firstly let \( z_{1}=x_{1}+i y_{1} \) and \( z_{2}=x_{2}+i y_{2} \)
Thus, \( z_{1} z_{2}=\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right) \)
\(\begin{array}{l}
=x_{1}\left(x_{2}+\mathrm{iy}_{2}\right)+\mathrm{iy}_{1}\left(x_{2}+\mathrm{iy}_{2}\right) \\
=x_{1} x_{2}+\mathrm{ix}_{1} y_{2}+\mathrm{iy}_{1} x_{2}+\mathrm{i}_{2} y_{1} y_{2} \\
=x_{1} x_{2}+\mathrm{ix}_{2} y_{2}+\mathrm{iy}_{1} x_{2}-y_{1} y_{2}\left(\mathrm{i}_{2}=-1\right) \\
=\left(x_{1} x_{2}-y_{1} y_{2}\right)+\mathrm{i}\left(x_{1} y_{2}+y_{1} x_{2}\right) \\
=\operatorname{Re}\left(z_{1} z_{2}\right)=x_{1} x_{2}-y_{1} y_{2} \\
\therefore \operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Re} \mathrm{Re}_{1} \operatorname{Rez}_{2}-\operatorname{Im} \mathrm{I}_{1} \operatorname{Im} z_{2}
\end{array}\)
Hence, proved

\(=\left\{\frac{1}{1-4 i}-\frac{2}{1+i}\right\}\left\{\frac{3-4 i}{5+i}\right\}=\left[\frac{(1+i)-2(1-4 i)}{(1-4 i)(1+i)}\right]\left[\frac{3-4 i}{5+i}\right]\)
\(=\left[\frac{1+i-2+8 i}{1+i-4 i-4 i^{2}}\right]\left[\frac{3-4 i}{5+i}\right]=\left[\frac{-1+9 i}{5-3 i}\right]\left[\frac{3-4 i}{5+i}\right]\)
\(=\left[\frac{-3+4 i+27 i-36 i^{2}}{25+5 i-15 i-3 i^{2}}\right]=\frac{33+31 i}{28-10 i}=\frac{33+31 i}{2(14-5 i)}\)
\( =\frac{(33+31 i)}{2(14-5 i)} \times \frac{14+5 i}{14+5 i} \) [on multiplying by numerator and denominator by
\( 14-5 i \)
\(=\frac{462+165 i+434 i+155 i^{2}}{2\left[(14)^{2}-(5 i)^{2}\right]}=\frac{307+533 i}{2\left(196-25 i^{2}\right)}\)
\(=\frac{307+599 i}{2(221)}=\frac{307+599 i}{442}=\frac{307}{442}+\frac{599 i}{442}\)
This is the required standard from.

\(\mathrm{x}-i \mathrm{y}=\sqrt{\frac{a-i b}{c-i d}}\)
\(=\sqrt{\frac{a-i b}{c-i d} \times \frac{c+i d}{c+i d}}[\text { on multiplying numerator and denominator by }(\mathrm{c}+\mathrm{id})]\)
\(=\sqrt{\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}}\)
\(=(x-i y)^{2}=\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}\)
\(=\mathrm{x}^{2}-\mathrm{y}^{2}-2 \mathrm{ixy}=\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}\)
On comparing real and imaginary parts, we obtain
\(\mathrm{X}_{2}-\mathrm{Y}_{2}=\frac{a c+b d}{c^{2}+d^{2}},-2 x y=\frac{a d-b c}{c^{2}+d^{2}}\)
\(\qquad\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}-4 x^{2} y^{2}\ldots(1)\)
\(\left\{\frac{a c+b d}{c^{2}+d^{2}}\right\}^{2}+\left\{\frac{a d-b c}{c^{2}+d^{2}}\right\}[\text { using }(1)]\)
\(=\frac{a^{2} c^{2}+b^{2} d^{2}+2 a c b d+a^{2} d^{2}+b^{2} c^{2}-2 a d b c}{\left(c^{2}+d^{2}\right)^{2}}\)
\(=\frac{a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}}{\left(c^{2}+d^{2}\right)^{2}}\)
\(=\frac{a^{2}\left(c^{2}+d^{2}\right)+b^{2}\left(c^{2}+d^{2}\right)}{\left(c^{2}+d^{2}\right)^{2}}\)
\(=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}\)
Hence, prove

Here, \( \mathrm{Z}=\frac{1+7 i}{(2-i)^{2}} \)
\(=\frac{1+7 i}{(2-i)^{2}}=\frac{1+7 i}{4+i^{2}-4 i}=\frac{1+7 i}{4-1-4 i}\)
\(=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}=\frac{3+4 i+21 i+28 i^{2}}{3^{2}+4^{2}}\)
\(=\frac{3+4 i+21 i-28}{25}=\frac{-25+25 i}{25}\)
\(=-1+i\)
Let \( r \cos \theta=-1 \) and \( r \sin \theta=1 \)
On squaring and adding, we obtain \( r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1 \)
\(=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2\)
\(=r^{2}=2\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]\)
\(=r=\sqrt{2}\)
\( \therefore \sqrt{2} \cos \theta=-1 \) and \( \sqrt{2} \sin \theta=1 \)
\( =\cos \theta=\frac{-1}{\sqrt{2}} \) and \( \sin \theta=\frac{1}{\sqrt{2}} \)
\( \therefore \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \)
[As \( \theta \) lies in II quadrant]
\( \therefore \mathrm{z}=\mathrm{r} \cos \theta+i \mathrm{r} \sin \theta \)
\( =\sqrt{2} \cos \frac{3 \pi}{4}+i \sqrt{2} \sin \frac{3 \pi}{4}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right) \)
This is the required polar form.

Here, \( \mathrm{z}=\frac{1+3 i}{1-2 i} \)
\(=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}\)
\(=\frac{1+2 i+3 i-6}{1+4}\)
\(=\frac{-5+5 i}{5}=-1+i\)
Let, \( r \cos \theta=-1 \) and \( r \sin \theta=1 \)
On squaring and adding, we obtain
\(r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1\)
\(=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2\)
\(=r^{2}=2\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]\)
\(=r=\sqrt{2}\)
\(\therefore \sqrt{2} \cos \theta=-1 \text { and } \sqrt{2} \sin \theta=1\)
\(=\cos \theta=\frac{-1}{\sqrt{2}} \text { and } \sin \theta=\frac{1}{\sqrt{2}}\)
\( \theta=\pi-\frac{3 \pi}{4}=\frac{3 \pi}{4} \) [As \( \theta \) lies in II quadrant]
\( \therefore \mathrm{z}=\mathrm{r} \cos \theta+i r \sin \theta=\sqrt{2} \cos \frac{3 \pi}{4}+i \sqrt{2} \sin \frac{3 \pi}{4}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right) \)
This is the required polar form.

We have, \( 9 x^{2}-12 x+20=0 \)
This equation can be rewritten as follows:
\(9 x^{2}-12 x+20=0\)
we have to compare this with \( \mathrm{a}x^2+\mathrm{b}x+\mathrm{c}=0 \) where \( \mathrm{a}=9, \mathrm{~b}=-12 \) and \( \mathrm{c}=20 \)
Thus, the discriminant of the above given equation is:
\(D=b^{2}-4 a c\)
\(=(-12)^{2}-4 \times 9 \times 20\)
\(=144-720\)
\(=-576\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(12) \pm \sqrt{-576}}{2 \times 9}=\frac{12 \pm \sqrt{576} i}{18} \quad[\sqrt{-1}=i]\)
\(=\frac{12 \pm 24 i}{18}=\frac{6(2 \pm 4 i)}{18}=\frac{2 \pm 4 i}{3}=\frac{2}{3} \pm \frac{4}{3} i\)

We have, \( 2 x^{2}-4 x+3=0 \)
This equation can be rewritten as follows:
\(2 x^{2}-4 x+3=0\)
we have to compare this with \( \mathrm{a}x^2+\mathrm{b}x+\mathrm{c}=0 \) where \( \mathrm{a}=2, \mathrm{~b}=-4 \) and \( \mathrm{c}=3 \)
Thus, the discriminant of the above given equation is:
\(D=b^{2}-4 a c\)
\(=(-4)^{2}-4 \times 2 \times 3\)
\(=16-24\)
\(=-8\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-4) \pm \sqrt{8}}{2 \times 2}=\frac{4 \pm 2 \sqrt{2} i}{4} \quad[\sqrt{-1}=i]\)
\(=\frac{2+\sqrt{2} i}{2}=1 \pm \frac{\sqrt{2}}{2} i\)

We have, \( 27 x^{2}-10 x+1=0 \)
we have to compare this with \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \) where \( \mathrm{a}=27, \mathrm{~b}=-10 \) and \( \mathrm{c}=1 \)
Thus, the discriminant of the above given equation is:
\(D=b^{2}-4 a c\)
\(=(-10)^{2}-4 \times 27 \times 1\)
\(=100-108\)
\(=-8\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-10) \pm \sqrt{-8}}{2 \times 27}=\frac{10 \pm 2 \sqrt{2} i}{54} \quad[\sqrt{-1}=i]\)
\(=\frac{5 \pm \sqrt{2} i}{27}=\frac{5}{27}+\frac{\sqrt{2}}{27} i\)

We have, \( 21 x^{2}-28 x+10=0 \)
we have to compare this with \( \mathrm{a}x^2+\mathrm{b}x+\mathrm{c}=0 \), where \( \mathrm{a}=21, \mathrm{~b}=-28 \) and \( \mathrm{c}=10 \)
Thus, the discriminant of the above given equation is:
\(D=b^{2}-4 a c\)
\(=(-28)^{2}-4 \times 21 \times 10\)
\(=784-840\)
\(=-56\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-28) \pm \sqrt{-56}}{2 \times 21}=\frac{28 \pm \sqrt{56} i}{42} \quad[\sqrt{-1}=i]\)
\(=\frac{28 \pm 2 \sqrt{14} i}{42}=\frac{28}{42} \pm \frac{2 \sqrt{14}}{42} i=\frac{2}{3}+\frac{\sqrt{14}}{21} i\)

\(z_{1}=2-i, z_{2}-1+i\)
\(\therefore\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=\left|\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\right|\)
\(\quad=\left|\frac{4}{2-2 i}\right|=\left|\frac{4}{2(1-i)}\right|\)
\(=\left|\frac{2}{1-i} \times \frac{1+i}{1+i}\right|=\left|\frac{2(1+i)}{\left(1^{2}-1^{2}\right)}\right|\)
\(=\left|\frac{2(1+i)}{1+1}\right| \quad\left[i^{2}=-1\right]\)
\(=\left|\frac{2(1+i)}{2}\right|\)
\(=|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2}\)
Thus, the value of \( \left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right| \) is \( \sqrt{2} \)

\(\mathrm{a}+i \mathrm{~b}=\frac{(x+i)^{2}}{\left(2 x^{2}+1\right)}\)
\(=\frac{x^{2}+i^{2}+2 x i}{2 x^{2}+1}\)
\(=\frac{x^{2}-1+2 x i}{2 x^{2}+1}\)
\(\quad=\frac{x^{2}-1}{2 x^{2}}+i \frac{2 x}{2 x^{2}}\)
On comparing real and imaginary parts, we obtain
\( \mathrm{a}=\frac{x^{2}-1}{2 x^{2}+1} \text { and } \mathrm{b}=\frac{2 x}{2 x^{2}+1}\)
\(\therefore a^{2}+b^{2}=\left\{\frac{x^{2}-1}{2 x^{2}+1}\right\}^{2}+\left\{\frac{2 x}{2 x^{2}+1}\right\}^{2}\)
\(=\frac{x^{4}+1-2 x^{2}+4 x^{2}}{(2 x+1)^{2}}\)
\(=\frac{x^{2}+1+2 x^{2}}{\left(2 x^{2}+1\right)^{2}}\)
\(=\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}\)
\(\therefore a^{2}+b^{2}=\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)}\)
Hence, proved

\(z_{1}=2-i, z_{2}=-2+i\)
\( z_{1} z_{2}=(2-i)(-2+i)=-4+2 i+2 i-i^{2}=-4+4 i- (-1)=-3+4 i \)
\(\overline{z_{1}} =2+i\)
\(\therefore \frac{z_{1} z_{2}}{\overline{z_{1}}}= \frac{-3+4 i}{2+i}\)
On multiplying numerator and denominator by (2-i), we obtain
\(\frac{z_{1} z_{2}}{\overline{z_{1}}}=\frac{(-3+4 i)(2-i)}{(2+i)(2-i)}=\frac{-6+3 i+8 i-4 i^{2}}{2^{2}+1^{2}}=\frac{-6+11 i-4(-1)}{2^{2} 1^{2}}\)
\(=\frac{-2+11 i}{5}=\frac{-2}{5}+\frac{11}{5} i\)
On comparing real parts, we obtain
\( \operatorname{Re} \frac{z_{1} z_{2}}{\overline{z_{1}}}=\frac{-2}{5} \)

\(z_{1}=2-i, z_{2}=-2+i\)
\( \frac{1}{z_{1} z_{2}}=\frac{1}{(2-i)(2+i)}=\frac{1}{(2)^{2}+(1)^{2}}=\frac{1}{5} \)
On comparing imaginary parts, we obtain
\(\operatorname{lm} \frac{1}{z_{1} z_{2}}=0\)

Let \( z=\frac{1+2 i}{1-3 i} \), then
\(\mathrm{z} =\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}=\frac{1+3 i+2 i+6 i^{2}}{1^{2}+3^{2}}=\frac{1+5 i+6(-1)}{1+9}\)
\(=\frac{-5+5 i}{10}=\frac{-5}{10}+\frac{5 i}{10}=\frac{-1}{2}+\frac{-i}{2}\)
\( =\text { Let } \mathrm{z}=\mathrm{r} \cos \theta+\mathrm{ir} \sin \theta\)
i.e., \( r \cos \theta=\frac{-1}{2} \) and \( r \sin \theta=\frac{1}{2} \)
On squaring and adding, we obtain
\(r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\left(\frac{-1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\)
\(\quad=r^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\quad=\mathrm{r}=\frac{1}{\sqrt{2}} \quad \text { [conventionally , } \mathrm{r} > 0 \text { ] }\)
\(\therefore \frac{1}{\sqrt{2}} \cos \theta=\frac{-1}{2} \text { and } \frac{1}{\sqrt{2}} \sin \theta=\frac{1}{2}\)
\(\quad=\cos \theta=\frac{-1}{\sqrt{2}} \text { and } \sin \theta=\frac{1}{\sqrt{2}}\)
\( \therefore \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \quad \) [As \( \theta \) lies in the II quadrant]
Therefore, the modules and argument of the given complex number are \( \frac{1}{\sqrt{2}} \) and \( \frac{3 \pi}{4} \) respectively.

Let \( \mathrm{z}=(x-i y)(3+5 i) \)
\(z=3 x+5 x i-3 y i-5 y i^{2}=3 x+5 x i-3 y i+5 y=(3 x+5 y)+ i(5 x-3 y)\)
\( \therefore \bar{z}=(3 x+5 y)-i(5 x-3 y)\)
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
\(3 x+5 y=-6 \ldots (1)\)
\(5 x-3 y=24 \ldots (2)\)
Multiplying equation (1) by 3 and (2) by 5 and then adding them, we obtain
\(9 x+15 y=18\)
\(25 x-15 y=120\)
\(34 x=102\)
\(\therefore x=\frac{102}{34}\)
\(x=3\)
Putting the value of \( x \) in equation (1), we obtain
\(3(3)+5 y=-6\)
\(=5 y=-6-9=-15\)
\(=y=3\)
Thus, the value of \( x \) and \( y \) are 3 and -3 respectively.
Hence, \( x=3 \) and \( y=-3 \)

\(\frac{1+i}{1-i}-\frac{1-i}{1+i}=\frac{(1+i)^{2}-(1-i)^{2}}{(1-i)(1+i)}\)
\(=\frac{1+i^{2}+2 i-1-i^{2}+2 i}{i^{2}+1^{2}}\)
\(\therefore \frac{4 i}{2}=2 i\)
\(\therefore\left|\frac{1+i}{1-i}-\frac{1+i}{1-i}\right|=|2 i|=\sqrt{2^{2}}=2\)

It is given in the question that,
\((x+i y)^{3}=u+i v\)
\(=x^{3}+(i y)^{3}+3 \cdot x \cdot i y(x+i y)=u+i v\)
\(=x^{3}+i^{3} y^{3}+3 x^{2} y i+3 x y^{2} i=u+i v\)
\(=x^{3}-4 y^{3}+3 x^{2} y i-3 x y^{2}=u+i v\)
\(=\left(x^{3}-3 x y^{2}\right)+i\left(3 x^{2} y-y^{3}\right)=u+i v\)
\(u=x^{3}-3 x y^{2}, v=3 x^{2} y-y^{3}\)
\(\therefore \frac{u}{x}+\frac{v}{y}=\frac{x^{3}-3 x y^{2}}{x}+\frac{3 x^{2} y-y^{3}}{y}\)
\(= \frac{x\left(x^{2}-3 y^{2}\right)}{x}+\frac{y\left(3 x^{2} y^{2}\right)}{y}\)
\(= x^{2}-3 y^{2}+3 x^{2}-y^{2}\)
\( =4\left(x^{2}-4 y^{2}\right)\)
\(\therefore \frac{u}{x}+\frac{v}{y}=4\left(x^{2}-y^{2}\right)\)
Hence, proved

Let \( \mathrm{a}=\mathrm{a}+\mathrm{i} \mathrm{b} \) and \( \beta=x+i y \)
It is given that, \( |\beta|=1 \)
\( \therefore \sqrt{x^{2}+y^{2}}=1\)
\(= x^{2}+y^{2}=1\)
\(\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=\left|\frac{(x+i y)-(a+i b)}{1-(a-i b)(x+i y)}\right|\)
\(=\left|\frac{(x-a)+i(y-b)}{1-(a x+a i y-i b x+b y)}\right|\)
\(=\left|\frac{(x-a)+i(y-b)}{(1-a x-b y)+i(b x-a y)}\right|\)
\(=\frac{|(x-a)+i(y-b)|}{|(1-a x-b y)+i(b x-a y)|}\)
\(=\frac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{(1-a x-b y)^{2}+(b x-a y)^{2}}}\)
\(=\frac{\sqrt{x^{2}+a^{2}-2 a x+y^{2}+b^{2}-2 b y}}{\sqrt{1+a^{2} x^{2}+b^{2} y^{2}-2 a x+2 a b x y-2 b y+b^{2} x^{2}+a^{2} y^{2}-2 a b x y}}\)
\(=\frac{\sqrt{\left(x^{2}+y^{2}\right)+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}\left(x^{2}+y^{2}\right)+b^{2}\left(y^{2}+x^{2}\right)-2 a x-2 b y}}\)
\(=\frac{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}}\)
\(=1\)
\(\therefore\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=1\)

\(|1-i|^{x}=2^{x}\)
\(=\left(\sqrt{1^{2}+(-1)^{2}}\right)^{x}=2^{x}\)
\(=(\sqrt{2})^{x}=2^{x}\)
\(=2^{\frac{x}{2}}=2^{x}\)
\(=x=2 x\)
\(=2 x-x=0\)
\(=x=0\)
Thus, 0 is the only integral solutions of the given equation.
Therefore, the number of non - integral solutions of \(Y\), e given equation is 0 .

\((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\)
\(\therefore|(a+i b)(c+i d)(e+i f)(g+i h)|=|A+i B|\)
\(=\sqrt{a^{2}+b^{2}} \times \sqrt{c^{2}+d^{2}} \times \sqrt{e^{2}+f^{2}} \times \sqrt{g^{2}+h^{2}}=\sqrt{A^{2}+B^{2}}\)
On squaring both sides, we obtain
\(\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=A^{2}+B^{2}\)
Hence, proved.

\(\left(\frac{1+i}{1-i}\right)^{n}=1\)
\(=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{n}=1\)
\(=\left(\frac{(1+i)^{2}}{1^{2}+i^{2}}\right)^{n}=1\)
\(=\left(\frac{1^{2}+i^{2}+2 i}{2}\right)^{n}=1\)
\(=\left(\frac{1-1+2 i}{2}\right)^{n}=1\)
\(=\left(\frac{2 i}{2}\right)^{n}=1\)
\(=i^{n}=1\)
\( \mathrm{m}=4 \mathrm{k} \), where k is some integer.
Therefore, the least positive integer is 1
Thus, the least positive integral value of \( m \) is \( 4(=4 \times 1) \).