Would you like to try solving a Korean high school math problem?

Hello, I am Jojo, a math teacher in Korea.

Today, I have brought some math problems that Korean high school students learn.

These problems are taught in the first year of high school and are about complex numbers.

If you know about complex numbers, you should be able to solve them.

I have prepared five problems.

Try solving them and compare your answers with the solutions provided below.

### Here is the first complex number problem.

#### i + i^2 + i^3 + \cdots + i^{2021} + i^{2022} = ?

Please calculate slowly and compare with the solution provided below.

korean math problem

\text{The powers of } i \text{ repeat every four terms:}

i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and repeat}

\text{Summing one complete cycle of four terms:}

i + (-1) + (-i) + 1 = 0

\text{Since the powers repeat every four terms, we can group the terms in cycles of four up to 2022:}

i + i^2 + i^3 + \cdots + i^{2021} + i^{2022}

\text{Recognize that 2022 can be divided into 505 complete cycles of four terms each,}

\text{plus the remaining two terms}

i^{2021} \text{ and } i^{2022}:

\left( \sum_{k=0}^{504} (i^{4k+1} + i^{4k+2} + i^{4k+3} + i^{4k+4}) \right) + i^{2021} + i^{2022}

\text{Each cycle sums to zero, so 505 cycles will also sum to zero:}

505 \times (i + (-1) + (-i) + 1) = 505 \times 0 = 0

\text{Now add the remaining terms } i^{2021} \text{ and } i^{2022}

i^{2021} = i, \quad i^{2022} = -1 \quad \Rightarrow \quad i – 1

\text{Therefore, the final sum is:}

i – 1

The second problem regarding complex numbers

#### The coefficient of the linear term in the polynomial (x – i)(x – i^2)(x – i^3) \cdots (x – i^{10}) is (a + bi).

#### Find the value of a + b where a and b are real numbers.

Take your time to think carefully and solve the problem yourself. Then, compare your solution with the one provided below.

korean math problem

To solve this problem, you don’t need to expand all 10 terms.

Instead, you should consider how the linear term is formed.

For example, with a polynomial with 2 terms, such as (x−1)(x−2), expansion is straightforward.

However, expanding all 10 terms is difficult.

The key idea is to focus on only the linear term that we need to find.

To do this, we’ll use the concept of “selection.” To form the linear term, 1 term out of the 10 must be x, and the remaining 9 terms will be complex numbers selected from the expansion.

forexample

(x {\color{blue}- i})(x {\color{blue} – i^2})(x {\color{blue} – i^3}) \cdots ({\color{red} x} – i^{10})

so, {\color{blue} (-i)(-i^2)(-i^3)(-i^4)(-i^5)(-i^6)(-i^7)(-i^8)(-i^9)} {\color{red}x}

In this way, we select terms to extract the coefficient of the linear term.

The coefficient of the linear term in the polynomial (x – i)(x – i^2)(x – i^3) \cdots (x – i^{10}) \text{ is:}

(-i)(-i^2)(-i^3)(-i^4)(-i^5)(-i^6)(-i^7)(-i^8)(-i^9) + (-i)(-i^2)(-i^3)(-i^4)(-i^5)(-i^6)(-i^7)(-i^8)(-i^{10}) + \cdots + (-i^2)(-i^3)(-i^4)(-i^5)(-i^6)(-i^7)(-i^8)(-i^9)(-i^{10})

= (-1)^9 i^{45} + (-1)^9 i^{46} + (-1)^9 i^{47} + \cdots + (-1)^9 i^{54}

= {(-i) + 1 + i + (-1)} + {(-i) + 1 + i + (-1)} + (-i) + 1

= 1 – i

\text{Therefore, } a = 1 \text{ and } b = -1, \text{ so:}

a + b = 1 + (-1) = 0

How was this problem?

Was it difficult due to the large number of terms to expand?

It wasn’t a problem where you need to expand everything directly.

The key was the idea behind it.

If you are familiar with the “binomial theorem,” you might have found it easier to understand.

Since the value we needed to find was the coefficient of the linear term, focusing on only expanding the linear term is crucial.

Now, shall we move on to the third problem?

### Third Complex Number Problem

\text{For a natural number } n,

\text{find the value of:}

\left(\cfrac{1 -i}{1 + i}\right)^{4n + 2} \left(\cfrac{1 + i}{1 -i}\right)^{4n}

\text{where } i = \sqrt{-1} .

Take your time to think and solve the problem, then compare your solution with the one provided below.

korean math problem

\text{To solve this problem, we need to simplify and evaluate the given expression:}

\left(\cfrac{1 -i}{1 + i}\right)^{4n + 2} + \left(\cfrac{1 + i}{1 -i}\right)^{4n}

First Fraction :

\cfrac{1 -i}{1 + i}

To simplify this, multiply the numerator and the denominator by the conjugate of the denominator:

\cfrac{1 -i}{1 + i} \times \cfrac{1 -i}{1 -i} = \cfrac{(1 -i)^2}{(1 + i)(1 -i)}

Calculate the numerator and denominator separately:

(1 -i)^2 = 1 -2i + i^2 = 1 -2i – 1 = -2i

(1 + i)(1 -i) = 1 -i^2 = 1 – (-1) = 2

so:

\cfrac{1 -i}{1 + i} = \cfrac{-2i}{2} = -i

Second Fraction:

\cfrac{1 + i}{1 -i}

Similarly, multiply the numerator and the denominator by the conjugate of the denominator:

\cfrac{1 + i}{1 -i} \times \cfrac{1 + i}{1 + i} = \cfrac{(1 + i)^2}{(1 -i)(1 + i)}

Calculate the numerator and denominator:

(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i – 1 = 2i

(1 -i)(1 + i) = 2

so:

\cfrac{1 + i}{1 -i} = \cfrac{2i}{2} = i

Evaluate the problem’s Expression

Now, substitute these results back into the original expression:

\left(-i\right)^{4n + 2} + \left(i\right)^{4n}

(-i)^{4n + 2} = [(-i)^2]^{2n + 1} = (-1)^{2n + 1} = -1

i^{4n} = (i^4)^n = 1^n = 1

Final Result

Add the two results:

\left(-i\right)^{4n + 2} + \left(i\right)^{4n} = -1 + 1 = 0

\text{The value of the given expression is } 0.

How did you find this problem? Were you able to solve it on your own?

It was a basic problem for Korean students. Now, shall we look at the fourth problem?

### The fourth problem concerning complex numbers.

\text{Given the equation }

i – \left( \cfrac{1 -i}{1 + i} \right)^{2013} = a + bi

\text{find the value of } a + b,

\text{where } a \text{ and } b \text{ are real numbers.}

\text{(Note: } i = \sqrt{-1}\text{)}

This problem is relatively easy!

I think you can definitely do it! Let’s solve it!

korean math problem

\text{Given the equation:}

i – \left( \cfrac{1 -i}{1 + i} \right)^{2013} = a + bi

\text{Simplify } \cfrac{1 -i}{1 + i}

\cfrac{1 -i}{1 + i}

\text{Multiply numerator and denominator by the conjugate of the denominator}

\cfrac{(1 -i)(1 -i)}{(1 + i)(1 -i)}

\text{Expand the numerator}

(1 -i)(1 -i) = 1 – 2i + i^2 = 1 – 2i – 1 = -2i

\text{Expand the denominator}

(1 + i)(1 -i) = 1 – i^2 = 1 – (-1) = 1 + 1 = 2

\text{Simplify the fraction}

\cfrac{1 -i}{1 + i} = \cfrac{-2i}{2} = -i

\text{Calculate } (-i)^{2013}

(-i)^{2013} = -i

\text{Substitute back into the original equation}

i – (-i) = i + i = 2i

\text{Express in the form } a + bi

a = 0, \, b = 2

\text{Find } a + b

a + b = 0 + 2 = 2

How was this problem?

It was easy, right?

Then let’s practice one more problem similar to the one we just solved for the last problem!

### The last problem concerning complex numbers

\text{Calculate the following expression:}

\left( \cfrac{1 + i}{1 – i} \right)^{100} + \left( \cfrac{1 – i}{1 + i} \right)^{999}

\text{(Note: } i = \sqrt{-1})

\text{Simplify } \cfrac{1 + i}{1 -i}

\text{Multiply numerator and denominator by the conjugate of the denominator}

\cfrac{1 + i}{1 -i} \times \cfrac{1 + i}{1 + i} = \cfrac{(1 + i)^2}{(1 -i)(1 + i)}

\text{Expand the numerator}

(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i – 1 = 2i

\text{Expand the denominator}

(1 – i)(1 + i) = 1 – i^2 = 1 – (-1) = 1 + 1 = 2

\text{Simplify}

\cfrac{1 + i}{1 -i} = \frac{2i}{2} = i

\text{Raise to the power of 100 }

\left( \cfrac{1 + i}{1 -i} \right)^{100} = i^{100}

i^4 = 1 \text{, so } i^{100} = (i^4)^{25} = 1^{25} = 1

\text{Simplify } \cfrac{1 -i}{1 +i}

\text{Calculate } \cfrac{1 -i}{1 + i}

\cfrac{1 -i}{1 + i}

\text{Multiply numerator and denominator by the conjugate of the denominator}

\cfrac{1 -i}{1 + i} \times \cfrac{1 -i}{1 -i}

\cfrac{(1 -i)^2}{(1 + i)(1 -i)}

\text{Expand the numerator}

(1 – i)^2 = 1 – 2i + i^2 = 1 – 2i – 1 = -2i

\text{ Simplify}

\cfrac{1 -i}{1 + i} = \cfrac{-2i}{2} = -i

\text{Raise to the power of 999}

\left( \cfrac{1 -i}{1 + i} \right)^{999} = (-i)^{999}

-i \text{ is equivalent to } i^3 \text{, so } (-i)^{999} = (i^3)^{999}

i^4 = 1 \text{, so } i^{2997} = (i^4)^{749} \cdot i^1 = 1^{749} \cdot i = i

\left( \cfrac{1 + i}{1 -i} \right)^{100} + \left( \cfrac{1 -i}{1 + i} \right)^{999} = 1 + i

\text{So, the final result is 1 + i}

How did you find this problem?

Did you solve it well?

It might have been challenging to use -i cubed to find the 999th power at the end.

Still, I believe you can do it if you think through it slowly!

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