# Test of Mathematics Solution Objective 398 - Complex Number and Binomial Theorem

This is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

## Problem

If $$a_0, a_1, \cdots, a_n$$ are real numbers such that $$(1+z)^n = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n$$ for all complex numbers z, then the value of $$(a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2$$ equals

(A) $$2^n$$ ; (B) $$a_0^2 + a_1^2 + \cdots + a_n^2$$ ; (C) $$2^{n^2}$$ ; (D) $$2n^2$$ ;

## Sequential Hints

(How to use this discussion: Do not read the entire solution at one go. First, read more on the Key Idea, then give the problem a try. Next, look into Step 1 and give it another try and so on.)

### Key Idea

This is the generic use case of Complex Number $$\iota =\sqrt {-1}$$ and binomial theorem.

### Step 1

Note that $$i^2 = -1$$. Also, geometrically speaking, i = (0,1). Hence adding (1,0) to i (=(0,1)) gives us the point (1, 1). Polar coordinate of this point is $$( \sqrt 2, \frac{pi}{4} )$$. Here is a picture:

Try the problem with this hint before looking into step 2. Remember, no one learnt mathematics by looking at solutions.

At Cheenta we are busy with Complex Number and Geometry module. Additionally I.S.I. Entrance Mock Test 1 is also active now.

Replace $$z$$ by $$i$$. We have $$(1+z)^n = (\sqrt 2 , \frac {\pi}{4} )^n = (2^{n/2}, \frac{n \cdot \pi }{4} )$$ on the left hand side.

Now, replace $$z$$ by $$i$$ on the right hand side.

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Replacing z by $$i$$ on the right hand side we have $$(2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 + a_1 i + a_2 i^2 + a_3 I^3 \cdots + a_n I^n$$. This implies $$(2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 - a_2 + a_4 - \cdots + i (a_1 - a_3 + a_5 - \cdots )$$

Think now, what the following expression represents: $$(a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2$$

It represents the square of the length of point $$(2^{n/2}, \frac{n \cdot \pi }{4} )$$. That is simply $$(2^{n/2})^2 = 2^n$$

Hence the answer is option A.

## More Resources

Look into the following notes on Complex Number and Geometry module.

# Test of Mathematics Solution Subjective 88 - Complex Numbers with a Property

This is a Test of Mathematics Solution Subjective 88 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem:

A pair of complex numbers $$z_1, z_2$$ is said to have the property $$P$$ if for every complex number $$z$$ we find real numbers $$r$$ and $$s$$ such that $$z=rz_1 + sz_2$$.Show that a pair of complex numbers has property $$P$$ if and only if the points $$z_1,z_2$$ and $$0$$ on the complex plane are not collinear.

## Solution:

Let the complex numbers $$z_1,z_2,0$$ be collinear, and the line joining them make an angle $$\theta$$ with the x-axis. This means that:

$$arg(z_1) =arg(z_2) = \theta$$

$$=> z_1 = |z_1| (cos\,\theta + i sin\, \theta)$$

Similarly,

$$=> z_2 = |z_2| (cos\,\theta + i sin\, \theta)$$

Therefore, $$z=rz_1 + sz_2$$

$$=> z =r |z_1| (cos\,\theta + i sin\, \theta) + s|z_2| (cos\,\theta + i sin\, \theta)$$

$$=> z =(r |z_1| + s|z_2|) (cos\,\theta + i sin\, \theta)$$

Which implies that $$z$$ lies on the same line that joins $$z_1$$ and $$z_2$$. But that is not true, as $$z$$ can be any complex number.

Thus the assumption that $$z_1, z_2, 0$$ are collinear is false.

Hence Proved.

# 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.

# 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.

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