Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

$-1+\sqrt{3} i$ divided by 2 is a cube root of unity. (It comes from x²+x+1=0)

When (1-i) has power 2 it becomes -2i

So what happens in power 6...

$(\sqrt[3]{1} )^6=1$ so the first one with of power 6 becomes 2⁶.

$1-i$ of power 6 becomes (-2i)³ or -2³.

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By division, we get -2³.

But this was power 6, so we need power 5 of -2³.

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So the answer will be -2¹⁵.

Answer 2

EXPLANATION.

$\sf \displaystyle \bigg(\frac{- 1 + \sqrt{3} \iota }{1 - \iota} \bigg)^{30}$

As we know that,

Concepts :

Exponential forms.

Let z be a complex number then the exponential form of complex number is given by, $z = r.e^{\iota \theta}$

Principal argument :

The value of θ such that - π < θ ≤ π is called principal argument. it is represented by amp(z).

Using this concepts in this question, we get.

⇒ - 1 + √3ι.

⇒ r = √(-1)² + (√3)².

⇒ r = √1 + 3.

⇒ r = √4.

⇒ r = 2.

⇒ amp(z) = - 1 + √3ι.

⇒ amp(z) = (√3)/(-1).

⇒ amp(z) = - √3.

⇒ amp(z) = - π/3.

As we can see that,

- 1 + √3ι lies on 2nd quadrant.

⇒ amp(z) = π - π/3.

⇒ amp(z) = 2π/3.

⇒ 1 - ι.

⇒ r = √(1)² + (1)².

⇒ r = √1 + 1.

⇒ r = √2.

⇒ amp(z) = (-1)/(1).

⇒ amp(z) = - π/4.

Now, we can write equation as,

$\sf \displaystyle \bigg[\frac{2e^{\iota \frac{2\pi}{3} } }{\sqrt{2} e^{- \iota \frac{\pi}{4} } } \bigg]^{30}$

$\sf \displaystyle \bigg[ \sqrt{2} e^{\iota (\frac{2\pi}{3} + \frac{\pi}{4} ) } \bigg]^{30}$

$\sf \displaystyle \bigg[\sqrt{2} e^{\iota (\frac{11 \pi}{12} )} \bigg]^{30}$

$\sf \displaystyle (\sqrt{2} )^{30} [e^{\iota (\frac{11 \pi}{12} )}]^{30}$

$\sf \displaystyle 2^{15} \bigg[e^{\iota \frac{55\pi}{2} } \bigg]$

$\sf \displaystyle 2^{15} \bigg[e^{\iota (\frac{55\pi}{2} - 28 \pi) }\bigg]$

$\sf \displaystyle 2^{15} \bigg[e^{\iota (\frac{- \pi}{2} )} \bigg] = - 2^{15} \iota$

$\sf \displaystyle \boxed{\bigg(\frac{- 1 + \sqrt{3} \iota }{1 - \iota} \bigg)^{30} = - 2^{15} \iota}$

Option [1] is correct answer.