Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

Amplitude and Complex numbers - AIME 1996


Let P be the product of the roots of \(z^{6}+z^{4}+z^{2}+1=0\) that have a positive imaginary part and suppose that P=r(costheta+isintheta) where \(0 \lt r\) and \(0 \leq \theta \lt 360\) find \(\theta\)

  • is 107
  • is 276
  • is 840
  • cannot be determined from the given information

Key Concepts


Equations

Complex Numbers

Integers

Check the Answer


Answer: is 276.

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

here\(z^{6}+z^{4}+z^{2}+1\)=\(z^{6}-z+z^{4}+z^{2}+z+1\)=\(z(z^{5}-1)+\frac{(z^{5}-1)}{(z-1)}\)=\(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\) then \(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\)=0

Second Hint

gives \(z^{5}=1 for z\neq 1\) gives \(z=cis 72,144,216,288\) and \(z^{2}-z+1=0 for z \neq 1\) gives z=\(\frac{1+-(-3)^\frac{1}{2}}{2}\)=\(cis60,300\) where cis\(\theta\)=cos\(\theta\)+isin\(\theta\)

Final Step

taking \(0 \lt theta \lt 180\) for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276.

.

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ISI MStat 2015 PSA Problem 18 | Complex Number

This is a beautiful problem from ISI MSTAT 2015 PSA problem 18 based on complex number. We provide sequential hints so that you can try.

Complex Number - ISI MStat Year 2015 PSA Question 18


The set of complex numbers $z$ satisfying the equation \( (3+7 i) z+(10-2 i) \bar{z}+100=0\) represents, in the complex plane

  • a straight line
  • a pair of intersecting straight lines
  • a point
  • a pair of distinct parallel straight lines

Key Concepts


Complex number representation

Straight line

Check the Answer


Answer: is a pair of intersecting straight lines

ISI MStat 2015 PSA Problem 18

Precollege Mathematics

Try with Hints


Simplify the Complex. Just Solve.

Let \(z = x+iy, \bar{z} = x-iy\) Then the given equation reduces to \((13x-9y+100)+i(5x-7y) = 0\).
Which implies \(13x-9y+100 = 0, 5x-7y = 0\).
They do intersect.(?)

Yes! they intersect and to get the point of intersection just use substitution . Hence it gives a pair of intersecting straight lines.

ISI MStat 2015 PSA Problem 18
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Problem on Complex plane | AIME I, 1988| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Complex Plane.

Problem on Complex Plane - AIME I, 1988


Let w_1,w_2,....,w_n be complex numbers. A line L in the complex plane is called a mean line for the points w_1,w_2,....w_n if L contains points (complex numbers) z_1,z_2, .....z_n such that \(\sum_{k=1}^{n}(z_{k}-w_{k})=0\) for the numbers \(w_1=32+170i, w_2=-7+64i, w_3=-9+200i, w_4=1+27i\) and \(w_5=-14+43i\), there is a unique mean line with y-intercept 3. Find the slope of this mean line.

  • is 107
  • is 163
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


Answer: is 163.

AIME I, 1988, Question 11

Elementary Algebra by Hall and Knight

Try with Hints


First hint

\(\sum_{k=1}^{5}w_k=3+504i\)

and \(\sum_{k-1}^{5}z_k=3+504i\)

Second Hint

taking the numbers in the form a+bi

\(\sum_{k=1}^{5}a_k=3\) and \(\sum_{k=1}^{5}b_k=504\)

Final Step

or, y=mx+3 where \(b_k=ma_k+3\) adding all 5 equations given for each k

or, 504=3m+15

or, m=163.

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Complex numbers and Sets | AIME I, 1990 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Complex Numbers and Sets.

Complex Numbers and Sets - AIME I, 1990


The sets A={z:\(z^{18}=1\)} and B={w:\(w^{48}=1\)} are both sets of complex roots with unity, the set C={zw: \(z \in A and w \in B\)} is also a set of complex roots of unity. How many distinct elements are in C?.

  • is 107
  • is 144
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Complex Numbers

Sets

Check the Answer


Answer: is 144.

AIME I, 1990, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

18th and 48th roots of 1 found by de Moivre's Theorem

=\(cis(\frac{2k_1\pi}{18})\) and \(cis(\frac{2k_2\pi}{48})\)

Second Hint

where \(k_1\), \(K_2\) are integers from 0 to 17 and 0 to 47 and \(cis \theta = cos \theta +i sin \theta\)

zw= \(cis(\frac{k_1\pi}{9}+\frac{k_2\pi}{24})=cis(\frac{8k_1\pi+3k_2\pi}{72})\)

Final Step

and since the trigonometric functions are periodic every period \({2\pi}\)

or, at (72)(2)=144 distinct elements in C.

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Complex roots and equations | AIME I, 1994 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Complex roots and equations.

Complex roots and equations - AIME I, 1994


\(x^{10}+(13x-1)^{10}=0\) has 10 complex roots \(r_1\), \(\overline{r_1}\), \(r_2\),\(\overline{r_2}\).\(r_3\),\(\overline{r_3}\),\(r_4\),\(\overline{r_4}\),\(r_5\),\(\overline{r_5}\) where complex conjugates are taken, find the values of \(\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}\)

  • is 107
  • is 850
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Complex Roots

Equation

Check the Answer


Answer: is 850.

AIME I, 1994, Question 13

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

here equation gives \({13-\frac{1}{x}}^{10}=(-1)\)

\(\Rightarrow \omega^{10}=(-1)\) for \(\omega=13-\frac{1}{x}\)

where \(\omega=e^{i(2n\pi+\pi)(\frac{1}{10})}\) for n integer

Second Hint

\(\Rightarrow \frac{1}{x}=13- {\omega}\)

\(\Rightarrow \frac{1}{(x)(\overline{x})}=(13-\omega)(13-\overline{\omega})\)

=\(170-13(\omega+\overline{\omega})\)

Final Step

adding over all terms \(\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}\)

=5(170)

=850.

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Function of Complex numbers | AIME I, 1999 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Function of Complex Numbers and Integers.

Function of Complex Numbers - AIME I, 1999


Let f(z) =(a+bi)z where a,b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin given that |a+bi|=8 and that \(b^{2}\)=\(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 259
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Integers

Complex Numbers

Check the Answer


Answer: is 259.

AIME I, 1999, Question 9

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

Let z=1+i f(1+i)=(a+bi)(1+i)=(a-b)+(a+b)i The image point must be equidistant from (1,1) and(0,0) then the image point lie on the line with slope -1 and which passes through \((\frac{1}{2},\frac{1}{2})\) that is x+y=1

Second Hint

putting x=(a-b) and y=(a+b) gives 2a=1 and \(a=\frac{1}{2}\)

Final Step

and \((\frac{1}{2})^{2} +b^{2}=8^{2}\) then \(b^{2}=\frac{255}{4}\) then 255+4=259.

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Equations and Complex numbers | AIME I, 2019 Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2019 based on Equations and Complex numbers.

Equations and Complex numbers - AIME 2019


For distinct complex numbers \(z_1,z_2,......,z_{673}\) the polynomial \((x-z_1)^{3}(x-z_2)^{3}.....(x-z_{673})^{3}\) can be expressed as \(x^{2019}+20x^{2018}+19x^{2017}+g(x)\), where g(x) is a polynomial with complex coefficients and with degree at most 2016. The value of \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\) can be expressed in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n

  • is 107
  • is 352
  • is 840
  • cannot be determined from the given information

Key Concepts


Equations

Complex Numbers

Integers

Check the Answer


Answer: is 352.

AIME, 2019, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

here \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\)=s=\((z_1z_2+z_1z_3+....z_1z_{673})+(z_2z_3+z_2z_4+...+z_2z_{673})\)

\(+.....+(z_{672}z_{673})\) here

P=\((x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2)...(x-z_{673})(x-z_{673})(x-z_{673})\)

Second Hint

with Vieta's formula,\(z_1+z_1+z_1+z_2+z_2+z_2+.....+z_{673}+z_{673}+z_{673}\)=-20 then \(z_1+z_2+.....+z_{673}=\frac{-20}{3}\) the first equation and \({z_1}^{2}+{z_1}^{2}+{z_1}^{2}+{z_1z_2}+{z_1z_2}+{z_1z_2}+.....\)=\(3({z_1}^{2}+{z_2}^{2}+.....+{z_{673}}^{2})\)+\(9({z_1z_2}+{z_1z_3}+....+{z_{672}z_{673}})\)=\(3({z_1}^{2}+{z_2}^{2}+.....+{z_{673}}^{2})\)+9s which is second equation

Final Step

here \((z_1+z_2+.....+z_{673})^{2}=\frac{400}{9}\) from second equation then \(({z_1}^{2}+{z_2}^{2}+.....+{z_{673}}^{2})+2({z_1z_2}+{z_1z_3}+....+{z_{672}z_{673}})=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+.....+{z_{673}}^{2})+2s=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+.....+{z_{673}}^{2})=\frac{400}{9}\)-2s then with second equation and with vieta s formula \(3(\frac{400}{9}-2s)+9s\)=19 then s=\(\frac{-343}{9}\) then |s|=\(\frac{343}{9}\) where 343 and 9 are relatively prime then 343+9=352.

.

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Complex Numbers and prime | AIME I, 2012 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Complex Numbers and prime.

Complex Numbers and primes - AIME 2012


The complex numbers z and w satisfy \(z^{13} = w\) \(w^{11} = z\) and the imaginary part of z is \(\sin{\frac{m\pi}{n}}\), for relatively prime positive integers m and n with m<n. Find n.

  • is 107
  • is 71
  • is 840
  • cannot be determined from the given information

Key Concepts


Complex Numbers

Algebra

Number Theory

Check the Answer


Answer: is 71.

AIME I, 2012, Question 6

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

Taking both given equations \((z^{13})^{11} = z\) gives \(z^{143} = z\) Then \(z^{142} = 1\)

Second Hint

Then by De Moivre's theorem, imaginary part of z will be of the form \(\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}}\) where \(k \in {1, 2, upto 70}\)

Final Step

71 is prime and n = 71.

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Complex Numbers and Triangles | AIME I, 2012 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on complex numbers and triangles.

Complex numbers and triangles - AIME I, 2012


Complex numbers a,b and c are zeros of a polynomial P(z)=\(z^{3}+qz+r\) and \(|a|^{2}+|b|^{2}+|c|^{2}\)=250, The points corresponding to a,b,.c in a complex plane are the vertices of right triangle with hypotenuse h, find \(h^{2}\).

  • is 107
  • is 375
  • is 840
  • cannot be determined from the given information

Key Concepts


Complex Numbers

Algebra

Triangles

Check the Answer


Answer: is 375.

AIME I, 2012, Question 14

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

here q ,r real a real b,c complex and conjugate pair x+iy,x-iy then a+b+c=0 gives a=-2x and by given condition a-x=y then y=-3x

Second Hint

\(|a|^{2}+|b|^{2}+|c|^{2}\)=250 then 24\(x^{2}\)=250

Final Step

h distance between b and c h=2y=-6x then \(h^{2}=36x^{2}\)=36\(\frac{250}{24}\)=375.

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Complex Numbers | AIME I, 2009 | Problem 2

Try this beautiful problem from AIME, 2009 based on complex numbers.

Complex Numbers - AIME, 2009


There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.

  • 101
  • 201
  • 301
  • 697

Key Concepts


Complex Numbers

Theory of equations

Polynomials

Check the Answer


Answer: 697.

AIME, 2009, Problem 2

Complex Numbers from A to Z by Titu Andreescue .

Try with Hints


First hint

Taking z=a+bi

Second hint

then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)

Final Step

then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$

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