# 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

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.

.

# 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

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.

# 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

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.

# 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

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.

# 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

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.

# 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

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.

# 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

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.

.

# 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

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.

# 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

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.

# 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

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