Complex Number MCQ Questions & Answers in Algebra | Maths

Consider the equation
$$\eqalign{ & {x^2} + 2x + 2 = 0 \cr & x = {\left( {\sqrt 2 {e^{i\frac{{3\pi }}{4}}}} \right)^{15}} + {\left( {\sqrt 2 {e^{ - i\frac{{3\pi }}{4}}}} \right)^{15}} \cr & {\text{Let }}\alpha = - 1 + i,\,\,\beta = - 1 - i \cr & {\alpha ^{15}} + {\beta ^{15}} = {\left( { - 1 + i} \right)^{15}} + {\left( { - 1 - i} \right)^{15}} \cr & = {\left( {\sqrt 2 {e^{i\frac{{3\pi }}{4}}}} \right)^{15}} + {\left( {\sqrt 2 {e^{ - i\frac{{3\pi }}{4}}}} \right)^{15}} \cr & = {\left( {\sqrt 2 } \right)^{15}}\left[ {{e^{\frac{{i45\pi }}{4}}} + {e^{\frac{{ - i45\pi }}{4}}}} \right] \cr & = {\left( {\sqrt 2 } \right)^{15}}.2\cos \frac{{45\pi }}{4} \cr & = {\left( {\sqrt 2 } \right)^{15}}.2\cos \frac{{3\pi }}{4} \cr & = \frac{{ - 2}}{{\sqrt 2 }}{\left( {\sqrt 2 } \right)^{15}} \cr & = - 2{\left( {\sqrt 2 } \right)^{14}} \cr & = - 256 \cr} $$

$$\eqalign{ & \frac{{1 + z}}{{1 + \overline z }} = \frac{{z\overline z + z}}{{1 + \overline z }}\left( {\because \,\,\left| z \right| = 1\, \Rightarrow \,{{\left| z \right|}^2} = 1\,\, \Rightarrow \,z\overline z = 1} \right) \cr & \frac{{1 + z}}{{1 + \overline z }} = \frac{{z\left( {\overline z + 1} \right)}}{{1 + \overline z }} = z. \cr} $$

$$\eqalign{ & {\alpha ^5} = 1 \cr & \therefore \,\,{\text{index}} = \left| {1 + \alpha + {\alpha ^2} + {\alpha ^3} - {\alpha ^4}} \right| \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| {1 + \alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4} - 2{\alpha ^4}} \right| \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| {\frac{{1 - {\alpha ^5}}}{{1 - \alpha }} - 2{\alpha ^4}} \right| \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| { - 2{\alpha ^4}} \right| \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{\left| \alpha \right|^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \times 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2. \cr} $$

$$S:\left| {z - 2 + i} \right|\, \geqslant \sqrt 5 $$     represents boundary and outer region of circle with center $$\left( {2, - 1} \right)$$  and radius $$\sqrt 5 $$
$${{z_0} \in S},$$     such that $$\frac{1}{{\left| {{z_0} - 1} \right|}}$$   is the maximum.
∴ $${\left| {{z_0} - 1} \right|}$$   is minimum
$${{z_0} \in S}$$   with $${\left| {{z_0} - 1} \right|}$$   as minimum will be a point on boundary of circle of region $$S$$ which lies on radius of this circle, which passes through (1, 0).
$$\therefore \,\,{z_0},1,2 - i$$   are collinear, or $$\left( {{x_0},{y_0}} \right),\left( {1,0} \right),\left( {2, - 1} \right)$$     are collinear.
∴ Using slopes of paralled lines,
$$\eqalign{ & \frac{{{y_0}}}{{{x_0} - 1}} = \frac{{ - 1}}{{2 - 1}} \cr & \Rightarrow \,\,{y_0} = 1 - {x_0} \cr} $$

$$\eqalign{ & {\text{Now, }}\frac{{4 - {z_0} - {{\overline z }_0}}}{{{z_0} - {{\overline z }_0} + 2i}} \cr & = \frac{{4 - \left( {{z_0} + {{\overline z }_0}} \right)}}{{\left( {{z_0} - {{\overline z }_0}} \right) + 2i}} \cr & = \frac{{4 - 2{x_0}}}{{2i{y_0} + 2i}} \cr & = \frac{{4 - 2{x_0}}}{{2i - 2{x_0}i + 2i}} \cr & = \frac{{2\left( {2 - {x_0}} \right)}}{{2\left( {2 - {x_0}} \right)}i} \cr & = \frac{1}{i} = - i \cr & \therefore \,\,Arg\left( {\frac{{4 - {z_0} - {{\overline z }_0}}}{{{z_0} - {{\overline z }_0} - 2i}}} \right) \cr & = Arg\left( { - i} \right) \cr & = \frac{{ - \pi }}{2} \cr} $$

$$\eqalign{ & \sum\limits_{n = 1}^{13} {\left[ {{i^n} + {i^{n + 1}}} \right]} = \sum\limits_{n = 1}^{13} {{i^n}\left[ {1 + i} \right]} \cr & = \left( {1 + i} \right)\left[ {i + {i^2} + {i^3}.....\,{i^{13}}} \right] = \frac{{\left( {1 + i} \right)}}{{\left( {1 - i} \right)}}i\left[ {1 - {i^{13}}} \right] \cr & = \frac{{\left( { - 1 + i} \right)\left( {1 - {i^{13}}} \right)}}{{\left( {1 - i} \right)}} = \frac{{ - 1 + {i^{13}} + i - {i^{14}}}}{{\left( {1 - i} \right)}} \cr & = \frac{{ - 1 + {{\left( {{i^2}} \right)}^6}.i + i - {{\left( {{i^2}} \right)}^7}}}{{\left( {1 - i} \right)}} = \frac{{2i + 2{i^2}}}{{1 - {i^2}}} = \left( {i - 1} \right) \cr} $$

$$\eqalign{ & z + \left| z \right| = 8 + 12i, \cr & \Rightarrow x + iy + \sqrt {{x^2} + {y^2}} = 8 + 12i \cr & \Rightarrow x + \sqrt {{x^2} + {y^2}} = 8\,\,\,\,\,.....\left( {\text{i}} \right)\,\,\,\& \,\,\,\,y = 12\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & \left( {x = - 5} \right){\text{So, }}z = - 5 + 12i \cr & \Rightarrow \left| z \right| = \sqrt {25 + 144} = 13 \cr & \Rightarrow \left| {{z^2}} \right| = {\left| z \right|^2} = 169 \cr} $$

$$\eqalign{ & \left| {{z_1} + {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right| = \left| {24 + 7i} \right| + 6 = 25 + 6 = 31 \cr & {\text{Also, }}\left| {{z_1} + {z_2}} \right| = \left| {{z_1} - \left( { - {z_2}} \right)} \right| \geqslant \left| {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right| \cr & \Rightarrow \left| {{z_1} + {z_2}} \right| \geqslant \left| {25 - 6} \right| = 19 \cr} $$
Hence, the least value of $$\left| {{z_1} + {z_2}} \right|$$  is 19 and the greatest value is 25.

We have,
$${\left( {1 + {\omega ^2} + 2\omega} \right)^{3n}} - {\left( {1 + \omega + 2{\omega ^2}} \right)^{3n}}$$
We know that, $$1 + \omega + {\omega ^2} = 0\,\,{\text{and}}\,\,{\omega ^3} = 1$$
∴ given expression is equal to
$$\eqalign{ & {\left( {2\omega - \omega } \right)^{3n}} - {\left( {2{\omega ^2} - {\omega ^2}} \right)^{3n}} \cr & = \,{\left( \omega \right)^{3n}} - {\left( {{\omega ^2}} \right)^{3n}} \cr & = {\left( {{\omega ^3}} \right)^n} - {\left( {{\omega ^3}} \right)^{2n}} \cr & = 1 - 1 \cr & = 0 \cr} $$

$$\eqalign{ & \left| {z - 1} \right| = \left| {z + 1} \right| \cr & \Rightarrow \,\,{\left( {x - 1} \right)^2} + {y^2} = {\left( {x + 1} \right)^2} + {y^2} \cr} $$
⇒ $$x = 0 ,$$  which represents a straight line.
Note $$\left| {\frac{{z - \alpha }}{{z - \beta }}} \right| = k,$$   where $$\alpha ,\beta $$  are complex constants, represents a straight line if $$k = 1.$$

$$\eqalign{ & {\text{Here given }}{x^3} - 1 = 0 \cr & {\text{has roots }}\alpha \,\& \,\beta {\text{ which are complex}} \cr & x = \alpha \,\& \,\alpha = \omega \cr & x = \beta \,\& \,\beta = {\omega ^2} \cr & {\text{Now, }}{\left( {1 + 2\alpha + \beta } \right)^3} - {\left( {3 + 3\alpha + 5\beta } \right)^3} \cr & = {\left( {1 + 2\omega + {\omega ^2}} \right)^3} - {\left( {3 + 3\omega + 5{\omega ^2}} \right)^3} \cr & = {\left( {1 + \omega + {\omega ^2} + \omega } \right)^3} - {\left( {3\left( {1 + \omega + {\omega ^2}} \right) + 2{\omega ^2}} \right)^3} \cr & {\text{As we know, }}1 + \omega + {\omega ^2} = 0 \cr & = {\left( {0 + \omega } \right)^3} - {\left( {3\left( 0 \right) + 2{\omega ^2}} \right)^3} \cr & = {\omega ^3} - 8{\omega ^6} \cr & = 1 - 8\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\omega ^3} = 1} \right) \cr & = - 7 \cr & {\text{So, }}{\left( {1 + 2\alpha + \beta } \right)^3} - {\left( {3 + 3\alpha + 5\beta } \right)^3} = - 7 \cr} $$