Recall that j² = -1.

So, -j⁷
= - (j²)³ (j¹)
= - (-1)³ j
= - (-1) j
= j

(-j)⁶
= (-1)⁶ j⁶
= j⁶
= (j²)³
= (-1)³
= -1

- sqrt (-j²)
= - sqrt [- (-1)]
= - sqrt 1
= -1

thanks! for qn6 i did it and got a positive 1 tho, what i did was to split(-j) ^6 into
(-j) ^2(-j)^2(-j)^2
= -(-1) x -(-1) x -(-1)
=1
why is this incorrect?

sorry for the late reply! i had alot of things going on in school the past few days

Because (-j)^2
= (-j) (-j)

Note: there are TWO negatives when you square the negative expression

= j^2
= -1

In short, -j^2 is not the same as (-j)^2

sorry i still dont really understand...

In mathematical operations of BODMAS,

- Brackets
- Order (power)
- Division/Multiplication

In -1^2, the power 2 acts first before the negative is applied, so -1^2 becomes -1.

In (-1)^2, the brackets take precedence, so (-1)^2 means (-1) times (-1) and the two negative signs disappear.

In your working, you wrote

(-j)^2

as

- (-1)

It’s actually -j^2 that’s equal to - (-1).

ohh i get it now, but if - j^2 = - (-1), what would (-j)^2 be written as? or j=-1 cannot be inserted into (-j)^2 at all and it has to be split up into (-1)^2(j)^2?

Technically, in me doing (-j)^2, I used the indices rule, (-j)^2 = (-1)^2 times j^2.

It’s this (-1)^2 which makes the negative sign in front of j^2 disappear.