Can you solve korean high school math Question?

Hello, I’m JoJo teacher.

I’m teaching math in Korea.

Today, I have a question related to complex numbers.

Would you like to give it a try?

[Can you solve korean high school math Question?]

I will explain this Question.

(1) $\; a_1, \; a_2, \; a_3, \cdots, \; a_{30} \;$ each have one of three possible values: $-1\;or\; i\;or\; 1+i$ .

(2) $\; (a_1)^2+(a_2)^2+(a_3)^2+\cdots +(a_{30})^2 = 5 + 14i$

What is the sum of the real and imaginary parts of $\; a_1+a_2+ \cdots +a_{30} \; ?$

$(i = \sqrt{-1})$

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I’ve interpreted the Question.

Do you understand what it means?

If you do, please give it a try.

Once you’ve solved it, you can refer to the explanation below.

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[solution]

$(-1)^2=1, \; i^2=-1, \; (1+i)^2=2i$

$(a_1)^2 + (a_2)^2 + (a_3)^2 + \cdots + (a_{30})^2 = 5 + 14i$

Since $14i$ is given, we can see that there are seven occurrences of $2i$ .

Now, we need to determine the number of occurrences of 1 and -1.

Let’s denote the number of occurrences of 1 as $x$ and the number of occurrences of -1 as $y$ .

Since $1+i$ appears $7$ times, $-1$ and $i$ should each appear 23 times.

$x + y = 23 \cdots A$

We have x occurrences of 1 and y occurrences of -1, and their sum should be equal to 5 when combined.

$x -y = 5 \cdots B$

We will solve the system of equations consisting of equation A and equation B.

$x = 14 , \; y = 9$

We can determine that $-1$ appears 14 times, $i$ appears 9 times, and $(1+i)$ appears 7 times.

finally

$-1\times14 + i \times 9 + (1+i)\times 7 = -14 + 9i + 7 + 7i = -7 + 16i$

$-7 + 16i \longrightarrow -7 + 16 = 9$

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How was the question today?

I know it was difficult, so it was challenging for me to explain as well.

However, if you take your time to think about it, you’ll figure it out.

Let’s think about how many times -1, i, and (1+i) were used in the question.

In addition to today’s question, I also have a question that I posted before.

Give it a try as well.

See you next time!

My youtube -> https://www.youtube.com/channel/UCnJ-GLzfJdWjs04eQoxfY7g

[Korea ver]

안녕하세요. 저는 한국에서 수학을 가르치고있는 조조쌤입니다.

오늘은 복소수 문제를 가지고왔습니다.

복소수에 대해서 잘 아시나요?

한번 풀어봅시다!

[문제]

문제지가 지저분해보이지만 잘 보이죠?

천천히 생각하면서 풀어보세요.

다 풀었다면 아래에 나와있는 해설과 비교해보세요.

[해설]

$(-1)^2=1, \; i^2=-1, \; (1+i)^2=2i$

$(a_1)^2 + (a_2)^2 + (a_3)^2 + \cdots + (a_{30})^2 = 5 + 14i$

$14i$ 라고 주어졌기 때문에 $2i$ 가 일곱개 있어야 한다는것을 알수있다.

이제 우리는 1과 -1이 몇개가 있는지를 생각해야합니다.

1의 개수를 x로, -1의 개수를 y로 표시해 봅시다.

(1 + i)가 7개 있어야한다는것을 알기 때문에 -1과 i는 23번 나타나야 합니다.

이것을 이용하여 식을 세우면

$x + y = 23 \cdots A$ 이렇게 됩니다.

그리고 1이 x개이고 -1이 y개이므로 , 둘의 합은 5여야 합니다.

$x -y = 5 \cdots B$

두개의 식으로 연립하여 풀면

$x = 14 , \; y = 9$ 입니다.

-1은 14개 있고, i는 9개 있다는 것을 알수있습니다.

따라서

$-1\times14 + i \times 9 + (1+i)\times 7 = -14 + 9i + 7 + 7i = -7 + 16i$

$-7 + 16i \longrightarrow -7 + 16 = 9$

${\color{red} 9}$

끝났습니다.

문제가 좀 어렵죠? 그래도 천천히 생각하면서 풀어본다면 충분히 풀었으리라고 생각됩니다.

그럼 다음시간에 재미있는 문제를 가지고 또 오겠습니다.

다음에 또봐요.