117839
If \(\mathbf{n}\) is integer which leaves remainder one when divided by three. Then \((1+\sqrt{3} \mathbf{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) equals
1 \(-2^{\mathrm{n}+1}\)
2 \(2^{\mathrm{n}+1}\)
3 \(-(-2)^{\mathrm{n}}\)
4 \(-2^{\mathrm{n}}\)
Explanation:
C Given, \((1+\sqrt{3} \mathrm{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) \(=\left[2\left(\frac{1+\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}+\left[2\left(\frac{1-\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}\) \(=\left(-2 \omega^2\right)^{\mathrm{n}}+(-2 \omega)^{\mathrm{n}}\) \(=(-2)^{\mathrm{n}}\left[\left(\omega^2\right)^{3 \mathrm{r}+1}+(\omega)^{3 \mathrm{r}+1}\right]\) \(\quad(\because \mathrm{n}=3 \mathrm{r}+1, \text { where r is an integer })\) \(=(-2)^{\mathrm{n}}\left(\omega^2+\omega\right)=-(-2)^{\mathrm{n}} \quad\left(\because \omega^2+\omega=-1\right)\)
VITEEE-2009
Complex Numbers and Quadratic Equation
117840
If \(\omega\) is an imaginary cube root of 1 , then the value of \(1(2-\omega)\left(2-\omega^2\right)+2(3-\omega)\left(3-\omega^2\right)+\ldots\) \(+(n-1)(n-\omega)\left(n-\omega^2\right)\) is
A Given \(z \text { is any complex number and }\) \(|z-1|=1\) \(\text { Let } z=x+\text { iy }\) \(\text { Then, } \mid(x+\text { iy })-1 \mid=1\) \(\sqrt{(x-1)^2+y^2}=1\) \(\text { Squaring both side }\) \((x-1)^2+y^2=1\) \(\text { From above equation it is clear that its represent a circle }\) \(\text { with centre (1,0) and radius } 1 \text {. }\) \(\text { imaginary axis }\) From above equation it is clear that its represent a circle with centre \((1,0)\) and radius 1 . Let \(T(z)\) is point on circle. Then \(\arg (z-1)=\theta\) \(\arg (z)=\frac{\theta}{2}\) \(\therefore \theta=2 \arg (\mathrm{z})\) From equation (i) and (ii), we get\(\arg (z-1)=2 \arg (z)\)
117839
If \(\mathbf{n}\) is integer which leaves remainder one when divided by three. Then \((1+\sqrt{3} \mathbf{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) equals
1 \(-2^{\mathrm{n}+1}\)
2 \(2^{\mathrm{n}+1}\)
3 \(-(-2)^{\mathrm{n}}\)
4 \(-2^{\mathrm{n}}\)
Explanation:
C Given, \((1+\sqrt{3} \mathrm{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) \(=\left[2\left(\frac{1+\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}+\left[2\left(\frac{1-\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}\) \(=\left(-2 \omega^2\right)^{\mathrm{n}}+(-2 \omega)^{\mathrm{n}}\) \(=(-2)^{\mathrm{n}}\left[\left(\omega^2\right)^{3 \mathrm{r}+1}+(\omega)^{3 \mathrm{r}+1}\right]\) \(\quad(\because \mathrm{n}=3 \mathrm{r}+1, \text { where r is an integer })\) \(=(-2)^{\mathrm{n}}\left(\omega^2+\omega\right)=-(-2)^{\mathrm{n}} \quad\left(\because \omega^2+\omega=-1\right)\)
VITEEE-2009
Complex Numbers and Quadratic Equation
117840
If \(\omega\) is an imaginary cube root of 1 , then the value of \(1(2-\omega)\left(2-\omega^2\right)+2(3-\omega)\left(3-\omega^2\right)+\ldots\) \(+(n-1)(n-\omega)\left(n-\omega^2\right)\) is
A Given \(z \text { is any complex number and }\) \(|z-1|=1\) \(\text { Let } z=x+\text { iy }\) \(\text { Then, } \mid(x+\text { iy })-1 \mid=1\) \(\sqrt{(x-1)^2+y^2}=1\) \(\text { Squaring both side }\) \((x-1)^2+y^2=1\) \(\text { From above equation it is clear that its represent a circle }\) \(\text { with centre (1,0) and radius } 1 \text {. }\) \(\text { imaginary axis }\) From above equation it is clear that its represent a circle with centre \((1,0)\) and radius 1 . Let \(T(z)\) is point on circle. Then \(\arg (z-1)=\theta\) \(\arg (z)=\frac{\theta}{2}\) \(\therefore \theta=2 \arg (\mathrm{z})\) From equation (i) and (ii), we get\(\arg (z-1)=2 \arg (z)\)
117839
If \(\mathbf{n}\) is integer which leaves remainder one when divided by three. Then \((1+\sqrt{3} \mathbf{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) equals
1 \(-2^{\mathrm{n}+1}\)
2 \(2^{\mathrm{n}+1}\)
3 \(-(-2)^{\mathrm{n}}\)
4 \(-2^{\mathrm{n}}\)
Explanation:
C Given, \((1+\sqrt{3} \mathrm{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) \(=\left[2\left(\frac{1+\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}+\left[2\left(\frac{1-\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}\) \(=\left(-2 \omega^2\right)^{\mathrm{n}}+(-2 \omega)^{\mathrm{n}}\) \(=(-2)^{\mathrm{n}}\left[\left(\omega^2\right)^{3 \mathrm{r}+1}+(\omega)^{3 \mathrm{r}+1}\right]\) \(\quad(\because \mathrm{n}=3 \mathrm{r}+1, \text { where r is an integer })\) \(=(-2)^{\mathrm{n}}\left(\omega^2+\omega\right)=-(-2)^{\mathrm{n}} \quad\left(\because \omega^2+\omega=-1\right)\)
VITEEE-2009
Complex Numbers and Quadratic Equation
117840
If \(\omega\) is an imaginary cube root of 1 , then the value of \(1(2-\omega)\left(2-\omega^2\right)+2(3-\omega)\left(3-\omega^2\right)+\ldots\) \(+(n-1)(n-\omega)\left(n-\omega^2\right)\) is
A Given \(z \text { is any complex number and }\) \(|z-1|=1\) \(\text { Let } z=x+\text { iy }\) \(\text { Then, } \mid(x+\text { iy })-1 \mid=1\) \(\sqrt{(x-1)^2+y^2}=1\) \(\text { Squaring both side }\) \((x-1)^2+y^2=1\) \(\text { From above equation it is clear that its represent a circle }\) \(\text { with centre (1,0) and radius } 1 \text {. }\) \(\text { imaginary axis }\) From above equation it is clear that its represent a circle with centre \((1,0)\) and radius 1 . Let \(T(z)\) is point on circle. Then \(\arg (z-1)=\theta\) \(\arg (z)=\frac{\theta}{2}\) \(\therefore \theta=2 \arg (\mathrm{z})\) From equation (i) and (ii), we get\(\arg (z-1)=2 \arg (z)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
117839
If \(\mathbf{n}\) is integer which leaves remainder one when divided by three. Then \((1+\sqrt{3} \mathbf{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) equals
1 \(-2^{\mathrm{n}+1}\)
2 \(2^{\mathrm{n}+1}\)
3 \(-(-2)^{\mathrm{n}}\)
4 \(-2^{\mathrm{n}}\)
Explanation:
C Given, \((1+\sqrt{3} \mathrm{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) \(=\left[2\left(\frac{1+\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}+\left[2\left(\frac{1-\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}\) \(=\left(-2 \omega^2\right)^{\mathrm{n}}+(-2 \omega)^{\mathrm{n}}\) \(=(-2)^{\mathrm{n}}\left[\left(\omega^2\right)^{3 \mathrm{r}+1}+(\omega)^{3 \mathrm{r}+1}\right]\) \(\quad(\because \mathrm{n}=3 \mathrm{r}+1, \text { where r is an integer })\) \(=(-2)^{\mathrm{n}}\left(\omega^2+\omega\right)=-(-2)^{\mathrm{n}} \quad\left(\because \omega^2+\omega=-1\right)\)
VITEEE-2009
Complex Numbers and Quadratic Equation
117840
If \(\omega\) is an imaginary cube root of 1 , then the value of \(1(2-\omega)\left(2-\omega^2\right)+2(3-\omega)\left(3-\omega^2\right)+\ldots\) \(+(n-1)(n-\omega)\left(n-\omega^2\right)\) is
A Given \(z \text { is any complex number and }\) \(|z-1|=1\) \(\text { Let } z=x+\text { iy }\) \(\text { Then, } \mid(x+\text { iy })-1 \mid=1\) \(\sqrt{(x-1)^2+y^2}=1\) \(\text { Squaring both side }\) \((x-1)^2+y^2=1\) \(\text { From above equation it is clear that its represent a circle }\) \(\text { with centre (1,0) and radius } 1 \text {. }\) \(\text { imaginary axis }\) From above equation it is clear that its represent a circle with centre \((1,0)\) and radius 1 . Let \(T(z)\) is point on circle. Then \(\arg (z-1)=\theta\) \(\arg (z)=\frac{\theta}{2}\) \(\therefore \theta=2 \arg (\mathrm{z})\) From equation (i) and (ii), we get\(\arg (z-1)=2 \arg (z)\)
117839
If \(\mathbf{n}\) is integer which leaves remainder one when divided by three. Then \((1+\sqrt{3} \mathbf{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) equals
1 \(-2^{\mathrm{n}+1}\)
2 \(2^{\mathrm{n}+1}\)
3 \(-(-2)^{\mathrm{n}}\)
4 \(-2^{\mathrm{n}}\)
Explanation:
C Given, \((1+\sqrt{3} \mathrm{i})^{\mathrm{n}}+(1-\sqrt{3} \mathrm{i})^{\mathrm{n}}\) \(=\left[2\left(\frac{1+\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}+\left[2\left(\frac{1-\sqrt{3} \mathrm{i}}{2}\right)\right]^{\mathrm{n}}\) \(=\left(-2 \omega^2\right)^{\mathrm{n}}+(-2 \omega)^{\mathrm{n}}\) \(=(-2)^{\mathrm{n}}\left[\left(\omega^2\right)^{3 \mathrm{r}+1}+(\omega)^{3 \mathrm{r}+1}\right]\) \(\quad(\because \mathrm{n}=3 \mathrm{r}+1, \text { where r is an integer })\) \(=(-2)^{\mathrm{n}}\left(\omega^2+\omega\right)=-(-2)^{\mathrm{n}} \quad\left(\because \omega^2+\omega=-1\right)\)
VITEEE-2009
Complex Numbers and Quadratic Equation
117840
If \(\omega\) is an imaginary cube root of 1 , then the value of \(1(2-\omega)\left(2-\omega^2\right)+2(3-\omega)\left(3-\omega^2\right)+\ldots\) \(+(n-1)(n-\omega)\left(n-\omega^2\right)\) is
A Given \(z \text { is any complex number and }\) \(|z-1|=1\) \(\text { Let } z=x+\text { iy }\) \(\text { Then, } \mid(x+\text { iy })-1 \mid=1\) \(\sqrt{(x-1)^2+y^2}=1\) \(\text { Squaring both side }\) \((x-1)^2+y^2=1\) \(\text { From above equation it is clear that its represent a circle }\) \(\text { with centre (1,0) and radius } 1 \text {. }\) \(\text { imaginary axis }\) From above equation it is clear that its represent a circle with centre \((1,0)\) and radius 1 . Let \(T(z)\) is point on circle. Then \(\arg (z-1)=\theta\) \(\arg (z)=\frac{\theta}{2}\) \(\therefore \theta=2 \arg (\mathrm{z})\) From equation (i) and (ii), we get\(\arg (z-1)=2 \arg (z)\)
