Question

show that root 3 upon 2 plus iota upon 2 cube is equal to iota
​

Answer 1

Answer:

Answer:

a²=b and b²=a proved.

Step-by-step explanation:

We are given that,

a=\frac{-1+i\sqrt{3}}{2}a=

2

−1+i

3

..... (1)

and, b= \frac{-1-i\sqrt{3}}{2}b=

2

−1−i

3

..... (2)

We have to prove that, a²=b and b²=a

Now, from (1) {Squaring both sides of (1)},

a²=\frac{1-3-i(2\sqrt{3})}{4}

4

1−3−i(2

3

)

⇒ a²= \frac{-2-i(2\sqrt{3})}{4}

4

−2−i(2

3

)

⇒ a²= \frac{-1-i\sqrt{3}}{2}

2

−1−i

3

⇒ a²= b {From equation (2)} (Proved)

Again, from (2) {Squaring both sides of (2)},

b²= \frac{1-3+i(2\sqrt{3})}{4}

4

1−3+i(2

3

)

⇒ b²= \frac{-2+i(2\sqrt{3})}{4}

4

−2+i(2

3

)

⇒ b²= \frac{-1+i\sqrt{3}}{2}

2

−1+i

3

⇒ b²= a {From equation (2)} (Hence, proved).