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# $Let{\text{ }}{{\text{z}}_1} = \dfrac{{2\sqrt 3 + i6\sqrt 7 }}{{6\sqrt 7 + i2\sqrt 3 }}{\text{ and }}{{\text{z}}_2} = \dfrac{{\sqrt {11} + i3\sqrt {13} }}{{3\sqrt {13} + i\sqrt {11} }}{\text{ then,}} \\ \left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}}} \right|{\text{ is equal to}} \\ \left( a \right){\text{ 47}} \\ \left( b \right){\text{ 264}} \\ \left( c \right){\text{ }}\left| {{z_1} - {z_2}} \right| \\ \left( d \right){\text{ }}\left| {{z_1} + {z_2}} \right| \\ \left( e \right){\text{ }}\left| {{z_1}{z_2}} \right| \\$

Last updated date: 13th Jul 2024
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$Rationali{\text{sing the denominator of g}}iven{\text{ }}{{\text{z}}_1} = \dfrac{{2\sqrt 3 + i6\sqrt 7 }}{{6\sqrt 7 + i2\sqrt 3 }} \\ We{\text{ get }}{{\text{z}}_1} = \dfrac{{2\sqrt 3 + i6\sqrt 7 }}{{6\sqrt 7 + i2\sqrt 3 }}{\text{ X }}\dfrac{{6\sqrt 7 - i2\sqrt 3 }}{{6\sqrt 7 - i2\sqrt 3 }} \\ This{\text{ is equal to }}\dfrac{{12\sqrt {21} + 12\sqrt {21} + 252i - 12i}}{{252 + 12}} \\ = {\text{ }}\dfrac{{24\sqrt {21} + 240i}}{{264}}{\text{ = }}\dfrac{{\sqrt {21} }}{{11}} + \dfrac{{10i}}{{11}} \\ Now{\text{ }}\dfrac{{\sqrt {21} }}{{11}} - \dfrac{{10i}}{{11}} = {\text{ }}\overline {{{\text{z}}_1}} \\ Now{\text{ doing this same rationalising with the given }}{{\text{z}}_2} = \dfrac{{\sqrt {11} + i3\sqrt 3 }}{{3\sqrt 3 - i\sqrt {11} }} \\ {\text{We get }}{{\text{z}}_2} = \dfrac{{\sqrt {11} + i3\sqrt 3 }}{{3\sqrt 3 - i\sqrt {11} }}{\text{ X }}\dfrac{{3\sqrt 3 + i\sqrt {11} }}{{3\sqrt 3 + i\sqrt {11} }}{\text{ }} \\ {\text{ = }}\dfrac{{3\sqrt {143} - 3\sqrt {143} + 117i + 11i}}{{117 + 11}} \\ We{\text{ get }}\dfrac{{128i}}{{128}} = i \\ So{\text{ our }}i = {\text{ }}\overline {{{\text{z}}_2}} \\ Now{\text{ we will find }}\dfrac{1}{{{z_1}}}{\text{ it is equal to }}\dfrac{{\overline {{z_1}} }}{{{{\left| {{z_1}} \right|}^2}}} \\ Hence{\text{ }}\dfrac{1}{{{z_1}}} = \dfrac{{\dfrac{{\sqrt {21} }}{{11}} - \dfrac{{10i}}{{11}}}}{{\sqrt {\dfrac{{21}}{{121}} + \dfrac{{100}}{{121}}} }}{\text{ = }}\dfrac{{\sqrt {21} }}{{11}} - \dfrac{{10i}}{{11}} = {\text{ }}\overline {{{\text{z}}_1}} {\text{ }} \\ {\text{Similarly }}\dfrac{1}{{{z_2}}}{\text{ = }}\dfrac{{\overline {{z_2}} }}{{{{\left| {{z_2}} \right|}^2}}}{\text{ = }}\dfrac{i}{{\sqrt 1 }} = i = {\text{ }}\overline {{{\text{z}}_2}} \\ Therefore\;\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_1}}} = \overline {{{\text{z}}_1}} + \overline {{{\text{z}}_2}} = \overline {{z_1} + {z_2}} \\ hence{\text{ }}\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_1}}}} \right|{\text{ = }}\left| {\overline {{z_1} + {z_2}} } \right| = \left| {{{\text{z}}_1} + {z_2}} \right|{\text{ as }}\left| {\overline z } \right|{\text{ = }}\left| z \right| \\ So{\text{ }}\left( d \right){\text{ option is the right answer}} \\ Note: {\text{ Whenever we encounter such problem we simply need to rationalise the}} \\ {\text{denominator of given complex numbers and eventually solving will take us to the right track}} \\ \\$