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# Complex Number- B.Stat. (Hons.) Admission Test 2005 – Objective Problem 4

## Competency in Focus: Complex Number

This problem from B.Stat 2005 Problem 4 - Objective Admission Test (Hons.)  is based on Demoivre's Theorem.

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The value of $\left\{\frac{1}{2}(-1+\sqrt{3} i)\right\}^{15}+\left\{\frac{1}{2}(-1-\sqrt{3} i)\right\}^{15}$ is
$\begin{array}{ll}{\text { (A) }-1 } & {\text { (B) } 0 \text { (C) } \frac{1}{2^{14}} \quad(\mathrm {D} ) 2}\end{array}$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.3.1" hover_enabled="0"]

B.Stat  2005 Problem 4 - Objective Admission Test (Hons.) [/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.2.2" inline_fonts="Abhaya Libre" open="on"]

### Complex Number

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]First of all know the Theorem that we are going to use. its called De-Moivres Theorem. See below $(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta$
for all integers $n$.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]I hope you must have solved it, becouse it just involves basic calculation and trignometric solutions. If not then see Next hint.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]$\left\{\frac{1}{2}(-1+\sqrt{3} i)\right\}^{15}+\left\{\frac{1}{2}(-1-\sqrt{3} i)\right\}^{15}$
$=\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \quad+\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)$
$=\cos \frac{2 \times 15 \times \pi}{3}+i \sin \frac{2 \times 15 \times \pi}{3}+\cos \frac{4 \times 15 \times \pi}{3}+i \sin \frac{4 \times 15 \times \pi}{3}$
$=\cos 10 \pi+i \sin 10 \pi+\cos 20 \pi+i \sin 20 \pi$
$=2$[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" custom_margin="20px||20px||false|false" global_module="50753" saved_tabs="all" locked="off"][et_pb_fullwidth_header title="I.S.I. & C.M.I. Program" button_one_text="Learn more" button_one_url="https://www.cheenta.com/isicmientrance/" header_image_url="https://www.cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="4.2.2" title_level="h2" title_font="Acme||||||||" background_color="#220e58" custom_button_one="on" button_one_text_color="#1a0052" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are: B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.
The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.