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4 years ago

Help I don’t know what to do

Mathematics

1 answer:

Allushta [10]4 years ago

6 0

Symbol i we call imaginary unit =>

This is basic

i^2= i * i = -1
i^3= i^2 * i = -1 * i = - i
i^4= i^3 * i = -i * i = - (i * i )=- (i^2)= - (-1) = 1
i^5= i^4 * i = 1 * i = i
i^6= i^5 * i = i * i = i^2 = -1
i^7= i^6 * i = -1 * i = - i
i^8= i^7 * i = -i * i = -(i * i)= -( i^2)= - (-1) = 1

You have all you need.
You conclude which simplifications of the power i are correct.

Good luck!!!!

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Well let's say that to compare these two numbers, we have to start with the definition first.

Definition

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Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

Definition II

$\displaystyle \large{ {a}^{2} = a \times a = |b| }$

Where b is the result from a×a. Let's see an example.

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$\displaystyle \large{ {2}^{2} = 2 \times 2 = 4} \\ \displaystyle \large{ {( - 2)}^{2} = ( - 2) \times ( - 2) = | - 4| = 4}$

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

$\displaystyle \large{ {y}^{2} = - 1}$

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

$\displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{ - 1} }$

Then where does plus-minus come from? It comes from one of Absolute Value propety.

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Solving absolute value always gives the plus-minus. Therefore...

$\displaystyle \large{ y = \pm \sqrt{ - 1} }$

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

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We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

Conclusion

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