Question

plz solve this 2 questions



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Answer 1

Answer:

1) Since w = cos 2π/3 + i sin 2π/3 , this means it's one of the cube root of unity and then we know the identity

1+w+w^2 = 0

Now, we'll modify the determinant a little bit, First transformation, adding C2 and C3 to C1

z+1+w+w^2 w. w^2

z+1+w+w^2 z+w^2 1

z+1+w+w^2. 1. z+w

=>. z. w. w^2.

z. z+w^2. 1

z. 1. z+w

Taking out z common from first column, and then solving the determinant along C1 we get,

z { (z+w^2)(z+w)-1) - z(z+w) + w^2 + w -w^2(z+w^2) }

= 0

=> z ( z^2 + wz + w^2 z + w^3 - z^2 - wz + w^2 + w - w^2z - w^4 ) = 0

=> z ( w^3 + w + w^2 - w) = 0

or, z(-w) = 0

or, z = 0

So, there's only one real root of the determinant equation that is z=0

2) By DeMoivre' s formula

(cis π/3)^3/4 = z

or,

z^4 = (cis π/3)^3 = cis π+2nπ = -1 +0i = -1

So,

z = (cis π+2nπ)^1/4 = cis (π/4+nπ/2)

=> z1 = cis π/4 , z2 = cis 3π/4 , z3 = cis 5π/4 , z4 = cis 7π/4

So, z1*z2*z3*z4

= e^iπ/4 * e^i3π/4 * e^i5π/4 * e^i7π/4

= e^i(4π) = 1

So, product of all these roots will be 1

Hope this helps you !

Answer 2

Question-1

We know,

• $\sf \omega = cos \:\dfrac{2\pi}{3}$

• $\sf 1+\omega+\omega^2 = 0$

Here,

We have add the transformations of determinant.

$\sf C_1\longrightarrow C_2+C_3$

$=\left[\begin{array}{ccc}\sf 1+z+\omega+\omega^2&\omega&\omega^2\\\sf1+z+\omega+\omega^2&z+\omega^2&1\\\sf1+z+\omega+\omega^2&1&z+\omega\end{array}\right]$

$=\left[\begin{array}{ccc}\sf z&\omega&\omega^2\\\sf z&z+\omega^2&1\\\sf z&1&z+\omega\end{array}\right]$

z is common in C₁ .

So,

solving the determinant along C₁ ,

$\implies\sf z [(z+w^2)(z+w)-1] - [z(z+w) + w^2] + [w -w^2(z+w^2)]=0$

$\implies\sf z ( z^2 + \omega z + \omega^2 z + \omega^3 - z^2 - \omegaz + \omega^2 + \omega - \omega^2z - \omega^4 ) = 0$

$\implies\sf z ( \omega^3 + \omega + \omega^2 - \omega) = 0$

$\implies\sf z(-\omega) = 0$

$\implies\sf z=0$

Therefore z = 0 is the only root.

$\boxed{\bold{\bf Number\:of\:distinct\:roots=1}}$

• Correct Option - Option 1

_________________________

Question-2

$\sf e^{i\theta}=cos\theta+i\:sin\theta$

$\sf (cos\dfrac{\pi}{3}+i\:sin\dfrac{\pi}{3})^{\frac{3}{4}}=(e^{i\theta})^{\frac{3}{4}}$

Here,

• $\sf \theta=\dfrac{\pi}{3}$

$\implies\sf \sf (cos\dfrac{\pi}{3}+i\:sin\dfrac{\pi}{3})^{\frac{3}{4}}=(e^{^{\frac{3}{4}}i\theta})$

$\implies\sf \sf (cos\dfrac{\pi}{3}+i\:sin\dfrac{\pi}{3})^{\frac{3}{4}}=(e^{{\frac{3}{4}}\times i\times\frac{\pi}{3}})$

$\implies\sf \sf (cos\dfrac{\pi}{3}+i\:sin\dfrac{\pi}{3})^{\frac{3}{4}}=(e^{\frac{\pi}{4}i})$

So,

$\sf cos\dfrac{\pi}{4}+i\:sin\dfrac{\pi}{4}=\sqrt{(cos\dfrac{\pi}{4})^2+(sin\dfrac{\pi}{4})^2$

$\implies\sf cos\dfrac{\pi}{4}+i\:sin\dfrac{\pi}{4}=\sqrt{(\dfrac{1}{\sqrt2})^2+(\dfrac{1}{\sqrt2})^2$

$\implies\sf cos\dfrac{\pi}{4}+i\:sin\dfrac{\pi}{4}=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}$

$\implies\sf cos\dfrac{\pi}{4}+i\:sin\dfrac{\pi}{4}=\sqrt{1}$

$\implies\sf cos\dfrac{\pi}{4}+i\:sin\dfrac{\pi}{4}=1$

Therefore,

$\boxed{\bold{\bf The\:product\:of\:all\:4\:values\:of\:(cos\dfrac{\pi}{3}+i\:sin\dfrac{\pi}{3})^{\frac{3}{4}}=1}}$

• Correct Option - Option 3