Complex Number MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {z^4} + {z^3} + {z^2} + {z^2} + z + 1 = 0 \cr & {\text{or, }}{z^2}\left( {{z^2} + z + 1} \right) + \left( {{z^2} + z + 1} \right) = 0 \cr & {\text{or, }}\left( {{z^2} + z + 1} \right)\left( {{z^2} + 1} \right) = 0 \cr & \therefore \,\,z = i, - i,\omega ,{\omega ^2}.\,{\text{For each, }}\left| z \right| = 1. \cr} $$

$$\eqalign{ & {\text{Since}}\,\left| {z - 1} \right| = 1 \cr & \therefore \,z - 1 = {e^{i\theta }},\,{\text{where}}\,\,\,\,\arg \left| {z - 1} \right| = \theta \cr & \therefore \,z = 1 + \cos \,\theta + i\,\sin \,\theta \cr & = \,2\cos \frac{\theta }{2}\left[ {\cos \frac{\theta }{2} + i\,\sin \,\frac{\theta }{2}} \right] \cr & = \,2\cos \frac{\theta }{2}.{e^{\frac{{i\theta }}{2}}} = 2{\cos ^2}\frac{\theta }{2} + 2i\,\sin \frac{\theta }{2}\cos \frac{\theta }{2} \cr} $$
Thus, $$arg\left( {z - 1} \right) = 2\arg z.$$

$$\eqalign{ & {\text{Given}}\,\left| {z - 4} \right| < \left| {z - 2} \right|{\text{Let}}\,z = x + iy \cr & \Rightarrow \,\,\left| {\left( {x - 4} \right) + iy)} \right| < \left| {\left( {x - 2} \right) + iy} \right| \cr & \Rightarrow \,\,{\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2} \cr & \Rightarrow \,\,{x^2} - 8x + 16 < {x^2} - 4x + 4 \cr & \Rightarrow \,\,12 < 4x \cr & \Rightarrow \,\,x > 3 \cr} $$
$$ \Rightarrow \,\,\operatorname{Re} \left( z \right) > 3$$

$$\eqalign{ & 1 - \sqrt 3 x + {x^2} = 0 \cr & x = \frac{{\sqrt 3 \pm i}}{2} \cr & x = \frac{{\sqrt 3 + i}}{2} = - i\overline w {\text{ and }}\overline x = \frac{{\sqrt 3 - i}}{2} = iw \cr & {\text{Now, }}{\left( {{x^n} - \frac{1}{{{x^n}}}} \right)^2} \cr & = {x^{2n}} + {\overline w ^{2n}} - 2 \cr & = - {\overline w ^{2n}} - {w^{2n}} - 2 \cr & = - 2 - \left[ {{{\overline w }^{2n}} + {w^{2n}}} \right] \cr & {\text{Now, }}\overline w = {w^2}{\text{ and }}1 + w + {w^2} = 0 \cr & {\text{Hence,}} \cr & - 2 - \left[ {{{\overline w }^{2n}} + {w^{2n}}} \right] \cr & = - 2 - \left[ {{w^{4n}} + {w^{2n}}} \right] \cr & = - 2 - \left[ { - 1} \right] \cr & = - 2 + 1 \cr & = - 1 \cr & {\text{Hence, }}\sum\limits_{n = 1}^{24} {{{\left( {{x^n} - \frac{1}{{{x^n}}}} \right)}^2}} = - 24 \cr} $$

Let the circle be $$\left| {z - {z_0}} \right| = r.$$  Then according to given conditions $$\left| {{z_0} - {z_1}} \right| = r + a\,\,{\text{and }}\left| {{z_0} - {z_2}} \right| = r + b.$$        Eliminating $$r,$$ we get $$\left| {{z_0} - {z_1}} \right| - \left| {{z_0} - {z_2}} \right| = a - b.$$
∴ Locus of center $${{z_0}}$$ is $$\left| {{z} - {z_1}} \right| - \left| {{z} - {z_2}} \right| = a - b,$$     which represents a hyperbola

$$\eqalign{ & \overrightarrow {OP} = \overrightarrow {OA} + \overrightarrow {AP} \cr & \Rightarrow \,\,\overrightarrow {OP} = \overrightarrow {OA} + \overrightarrow {OB} \cr & \Rightarrow \,\,\overrightarrow {OP} = 3{e^{i\frac{\pi }{4}}} + 4{e^{i\left( {\frac{\pi }{2} + \frac{\pi }{4}} \right)}} \cr & = 3{e^{i\frac{\pi }{4}}} + 4{e^{i\frac{\pi }{2}}}.{e^{i\frac{\pi }{4}}} \cr & = 3{e^{i\frac{\pi }{4}}} + 4i{e^{i\frac{\pi }{4}}} = {e^{i\frac{\pi }{4}}}\left( {3 + 4i} \right). \cr} $$

$$\left( {z + 1} \right)\left( {{z^2} + z + 2} \right) = 0;$$     non-real complex roots are found from $${{z^2} + z + 2 = 0}.$$

$$\eqalign{ & \sum\limits_{k = 1}^{10} {\left( {\sin \frac{{2k\pi }}{{11}} + i\cos \frac{{2k\pi }}{{11}}} \right)} \cr & = i\sum\limits_{k = 1}^{10} {\left( {\cos \frac{{2k\pi }}{{11}} - i\sin \frac{{2k\pi }}{{11}}} \right)} \cr & = i\sum\limits_{k = 1}^{10} {{e^{ - \frac{{2k\pi }}{{11}}i}}} \cr & = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - \frac{{2k\pi }}{{11}}i}} - 1} } \right\} \cr & = i\left[ {1 + {e^{ - \frac{{2\pi }}{{11}}i}} + {e^{ - \frac{{4\pi }}{{11}}i}} + ..... + 11\,{\text{terms}}} \right] - i \cr & = i\left[ {\frac{{1 - {{\left( {{e^{ - \frac{{2\pi }}{{11}}}}} \right)}^{11}}}}{{1 - {e^{ - \frac{{2\pi }}{{11}}i}}}}} \right] - i \cr & = i\left[ {\frac{{1 - {e^{ - 2\pi i}}}}{{1 - {e^{ - \frac{{2\pi }}{{11}}i}}}}} \right] - i \cr & = i \times 0 - i\,\,\,\,\,\,\,\,\,\,\,\left[ {\because {e^{ - 2\pi i}} = 1} \right] = - i \cr} $$

$$\eqalign{ & {z^{\frac{1}{3}}} = a - ib \cr & \Rightarrow z = {\left( {a - ib} \right)^3} \cr & \therefore x + iy = {a^3} + i{b^3} - 3i{a^2}b - 3a{b^2}. \cr & {\text{Then, }}x = {a^3} - 3a{b^2} \cr & \Rightarrow \frac{x}{a} = {a^2} - 3{b^2} \cr & y = {b^3} - 3{a^2}b \cr & \Rightarrow \frac{y}{b} = {b^2} - 3{a^2} \cr & {\text{So, }}\frac{x}{a} - \frac{y}{b} = 4\left( {{a^2} - {b^2}} \right) \cr} $$

$$\eqalign{ & {\log _{\frac{1}{2}}}\frac{{{{\left| z \right|}^2} + 2\left| z \right| + 4}}{{2{{\left| z \right|}^2} + 1}} < 0 = {\log _{\frac{1}{2}}}1 \cr & \Rightarrow \,\,\frac{{{{\left| z \right|}^2} + 2\left| z \right| + 4}}{{2{{\left| z \right|}^2} + 1}} > 1 \cr & {\text{or, }}{\left| z \right|^2} + 2\left| z \right| + 4 > 2{\left| z \right|^2} + 1 \cr & {\text{or, }}{\left| z \right|^2} - 2\left| z \right| - 3 < 0 \cr & {\text{or, }}\left( {\left| z \right| + 1} \right)\left( {\left| z \right| - 3} \right) < 0 \cr & \therefore \,\,\left| z \right| - 3 < 0. \cr} $$