Question

If a=-1+root3i/2,b=-1-root3i/2 then show that a²=b and b²=a.​

Answer 1

Hence we have show that  $a^2=b$ and $b^2=a$

Step-by-step explanation:

Given that $a=\frac{-1+\sqrt{3}i}{2}$ and $b=\frac{-1-\sqrt{3}i}{2}$

To show that $a^2=b$ and $b^2=a$

• Let us first show that $a^2=b$

Take LHS $a^2$

Substitute the value of a in the above expression we get

$=(\frac{-1+\sqrt{3}i}{2})^2$

$=\frac{(-1+\sqrt{3}i)^2}{2^2}$ ( by using the property $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

$=\frac{(-1)^2+(\sqrt{3}i)^2+2(-1)(\sqrt{3}i)}{4}$ ( by using the identity $(a+b)^2=a^2+b^2+2ab$ here a=-1 and b=$\sqrt{3}i$ )

$=\frac{1+3i^2-2i\sqrt{3}}{4}$

$=\frac{1+3(-1)-2i\sqrt{3}}{4}$ ( by using $i^2=-1$ )

$=\frac{1-3-2i\sqrt{3}}{4}$

$=\frac{-2-2i\sqrt{3}}{4}$

$=\frac{2(-1-i\sqrt{3}}{4}$

$=\frac{-1-\sqrt{3}i}{2}$

$=b=RHS$

Therefore $a^2=b$ hence proved

• Let us show that $b^2=a$

Take LHS $b^2$

Substitute the value of a in the above expression we get

$=(\frac{-1-\sqrt{3}i}{2})^2$

$=\frac{(-1-\sqrt{3}i)^2}{2^2}$ ( by using the property $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

$=\frac{(-1)^2+(-\sqrt{3}i)^2+2(-1)(-\sqrt{3}i)}{4}$ ( by using the identity $(a+b)^2=a^2+b^2+2ab$ here a=-1 and b=$-\sqrt{3}i$ )

$=\frac{1+3i^2+2i\sqrt{3}}{4}$

$=\frac{1+3(-1)+2i\sqrt{3}}{4}$ ( by using $i^2=-1$ )

$=\frac{1-3+2i\sqrt{3}}{4}$

$=\frac{-2+2i\sqrt{3}}{4}$

$=\frac{2(-1+i\sqrt{3}}{4}$

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

$=a=RHS$

Therefore $b^2=a$ hence proved.