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Maths 3 Marks Question

Complex Numbers and Quadratic Equations · CBSE STD 11 Science (CBSE - English Medium). 11 questions.

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Question 13 Marks
$z_1$ and $z_2$ are two complex numbers such that $|z_1| = |z_2|$ and $\text{arg(z}_1)+\text{arg(z}_2)=\pi,$ then show that $\text{z}_1=-\bar{\text{z}}_2$
Answer
Let $\text{z}_1=|\text{z}_1|(\cos\theta_1+\text{i}\sin\theta_1)$ and $\text{z}_2=|\text{z}_2|(\cos\theta_2+\text{i}\sin\theta_2)$
Given that, $|\text{z}_1|=|\text{z}_2|$
And arg $(\text{z}_1)+\text{arg}(\text{z}_2)=\pi$
$\Rightarrow\theta_1+\theta_2=\pi$
$\Rightarrow\theta_1=\pi-\theta_2$
Now, $\text{z}_1=|\text{z}_2|\big[\cos(\pi-\theta_2)+\text{i}\sin(\pi-\theta_2)\big]$
$\Rightarrow\text{z}_1=|\text{z}_2|(-\cos\theta_2+\text{i}\sin\theta_2)$
$\Rightarrow\text{z}_1=-|\text{z}_2|(\cos\theta_2-\text{i}\sin\theta_2)$
$\Rightarrow\text{z}_1=-\big[|\text{z}_2|(\cos\theta_2-\text{i}\sin\theta_2)\big]$
$\Rightarrow\text{z}_1=-\bar{\text{z}}_2$
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Question 23 Marks
If the real part of $\frac{\bar{\text{z}}+2}{\bar{\text{z}}-1}$ is 4, then show that the locus of the point representing z in the complex plane is a circle.
Answer
Let $\text{z}=\text{x}+\text{iy}$
Now, $\frac{\bar{\text{z}}+2}{\bar{\text{z}}-1}=\frac{\text{x}-\text{iy}+2}{\text{x}-\text{iy}-1}=\frac{\big[(\text{x}+2)-\text{iy}\big]\big[(\text{x}-1)+\text{iy}\big]}{\big[(\text{x}-1)-\text{iy}\big]\big[(\text{x}-1)+\text{iy}\big]}$
$=\frac{(\text{x}-1)(\text{x}+2)+\text{y}^2+\text{i}\big[(\text{x}+2)\text{y}-(\text{x}-1)\text{y}\big]}{(\text{x}-1)^2+\text{y}^2}$
Given that real part is 4
$\Rightarrow\frac{(\text{x}-1)(\text{x}+2)+\text{y}^2}{(\text{x}-1)^2+\text{y}^2}=4$
$\Rightarrow\text{x}^2+\text{x}-2+\text{y}^2=4(\text{x}^2-2\text{x}+1+\text{y}^2)$
$\Rightarrow3\text{x}^2+3\text{y}^2-9\text{x}+6=0,$ which represents a circle.
Hence, locus of z is circle.
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Question 33 Marks
Show that $\Big|\frac{\text{z}-2}{\text{z}-3}\Big|=2$ represents a circle. Find its center and radius.
Answer
We have $\Big|\frac{\text{z}-2}{\text{z}-3}\Big|=2$
Putting z = x + iy, we get
$\Big|\frac{\text{x}+\text{iy}-2}{\text{x}+\text{iy}-3}\Big|=2$
$\Rightarrow|\text{x}-2+\text{iy}|=2|\text{x}-3+\text{iy}|$
$\Rightarrow\sqrt{(\text{x}-2)^2+\text{y}^2}=2\sqrt{(\text{x}-3)^2+\text{y}^2}$
$\Rightarrow\text{x}^2-4\text{x}+4+\text{y}^2=4\big(\text{x}^2-6\text{x}+9+\text{y}^2\big)$
$\Rightarrow3\text{x}^2+3\text{y}^2-20\text{x}+32=0$
$\Rightarrow\text{x}^2+\text{y}^2-\frac{20}{3}\text{x}+\frac{32}{3}=0$
$\Rightarrow\Big(\text{x}-\frac{10}{3}\Big)^2+\text{y}^2+\frac{32}{3}-\frac{100}{9}=0$
$\Rightarrow\Big(\text{x}-\frac{10}{3}\Big)^2+(\text{y}-0)^2=\frac{4}{9}$
Hence, centre of the circle is $\Big(\frac{10}{3},0\Big)$ and radius is $\frac{2}{3}$
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Question 43 Marks
If $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy},$ what is the value of $x^2 + y^2$ ?
Answer
Given that$, \frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy}\ ...(\text{i})$
Taking conjugate on both sides$,$
$\frac{(\text{a}^2+1)^2}{2\text{a}+\text{i}}=\text{x}-\text{iy}\ ...(\text{ii})$
Multiplying $eq. (i)$ and $(ii)$ we have
$\frac{(\text{a}^2+1)^2(\text{a}^2+1)^2}{(2\text{a}-\text{i})(2\text{a}+\text{i})}=\text{x}^2+\text{y}^2$
$\Rightarrow\frac{(\text{a}^2+1)^4}{4\text{a}^2-\text{i}^2}=\text{x}^2+\text{y}^2$
$\Rightarrow\frac{(\text{a}^2+1)^4}{4\text{a}^2-\text{i}^2}=\text{x}^2+\text{y}^2$
Hence$,$ the value of $\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
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Question 53 Marks
Evaluate $\sum\limits^{13}_{\text{n}=1}\big(\text{i}^{\text{n}}+\text{i}^{\text{n}+1}\big),$ where $\text{n}\in\text{N}.$
Answer
$\sum\limits^{13}_{\text{n}=1}\big(\text{i}^{\text{n}}+\text{i}^{\text{n}+1}\big)=\sum\limits^{13}_{\text{n}=1}(1+\text{i})\text{i}^{\text{n}}$
$=(1+\text{i})(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9+\text{i}^{10}+\text{i}^{11}+\text{i}^{12}+\text{i}^{13})$
$=(1+\text{i})(\text{i}-1-\text{i}+1+\text{i}-1-\text{i}+1+\text{i}-1-\text{i}+1+\text{i})$
$=(1+\text{i})\text{i}=\text{i}+\text{i}^2=\text{i}-1$

Alternative Answer
$\sum\limits^{13}_{\text{n}=1}\big(\text{i}^{\text{n}}+\text{i}^{\text{n}+1}\big)=\sum\limits^{13}_{\text{n}=1}(1+\text{i})\text{i}^{\text{n}}$
$=(1+\text{i})(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9+\text{i}^{10}+\text{i}^{11}+\text{i}^{12}+\text{i}^{13})$
$=(1+\text{i})\frac{\text{i}(\text{i}^{{13}}-1)}{\text{i}-1}=(1+\text{i})\frac{\text{i}(\text{i}-1)}{\text{i}-1}=(1+\text{i})\text{i}=\text{i}+\text{i}^2=\text{i}-1$

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Question 63 Marks
If z and w are two complex numbers such that zw =1 and $\text{arg(z)}-\text{arg(w)}=\frac{\pi}{2},$ then show that $\bar{\text{z}}\text{w}=-\text{i}$
Answer
Let $\text{z}=|\text{z}|(\cos\theta_1+\text{i}\sin\theta_1)$ and $\text{w}=|\text{w}|(\cos\theta_2+\text{i}\sin\theta_2)$
Given that, $|\text{zw}|=|\text{z}||\text{w}|=1$
Also, $\arg(\text{z})-\arg(\text{w})=\frac{\pi}{2}$
$\theta_1-\theta_2=\frac{\pi}{2}$
Now, $\bar{\text{z}}\text{w}=|\text{z}|(\cos\theta_1+\text{i}\sin\theta_1)|\text{w}|(\cos\theta_2+\text{i}\sin\theta_2)$
$=|\text{z}||\text{w}|(\cos(-\theta_1)+\text{i}\sin(-\theta_1))(\cos\theta_2+\text{i}\sin\theta_2)$
$=1\big[\cos(\theta_2-\theta_1)+\text{i}\sin(\theta_2-\theta_1)\big]$
$=\Big[\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)\Big]$
$=1\big[0-\text{i}\big]$
$=\text{i}$
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Question 73 Marks
If $\frac{(1+\text{i})^2}{2-\text{i}}=\text{x}+\text{iy},$ then find the value of x + y.
Answer
Given that, $\frac{(1+\text{i})^2}{2-\text{i}}=\text{x}+\text{iy}$
$\Rightarrow\frac{1+\text{i}^2+2\text{i}}{2-\text{i}}=\text{x}+\text{iy}$
$\Rightarrow\frac{1-1+2\text{i}}{2-\text{i}}=\text{x}+\text{iy}$
$\Rightarrow\frac{2\text{i}}{2-\text{i}}=\text{x}+\text{iy}$
$\Rightarrow\frac{2\text{i}(2+\text{i})}{(2-\text{i})(2+\text{i})}=\text{x}+\text{iy}$
$\Rightarrow\frac{4\text{i}+2\text{i}^2}{4-\text{i}^2}=\text{x}+\text{iy}$
$\Rightarrow\frac{4\text{i}-2}{4+1}=\text{x}+\text{iy}$ $[\because\text{ i}^2=-1\big]$
$\Rightarrow\frac{-2+4\text{i}}{5}=\text{x}+\text{y}$
$=\frac{-2}{5}+\frac{4}{5}\text{i}=\text{x}+\text{iy}$
Comparing the real and imaginary parts, we get
$\text{x}=\frac{-2}{5}$ and $\text{y}=\frac{-4}{5}$
Hence, $\text{x}+\text{y}=\frac{-2}{5}+\frac{4}{5}=\frac{2}{5}$
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Question 83 Marks
What is the conjugate of $\frac{2-\text{i}}{(1-2\text{i})^2}$
Answer
We have $\text{z}=\frac{2-\text{i}}{(1-2\text{i})^2}$
$\text{z}=\frac{2-\text{i}}{1+4\text{i}^2-4\text{i}}=\frac{2-\text{i}}{1-4-4\text{i}}=\frac{2-\text{i}}{-3-4\text{i}}$
$=\frac{(2-\text{i})}{-(3+4\text{i})}=-\Big[\frac{(2-\text{i})(3-4)}{(3+4\text{i})(3-4\text{i})}\Big]$
$=-\Big(\frac{6-8\text{i}-3\text{i}+4\text{i}^2}{9+16}\Big)=-\frac{(-11\text{i}+2)}{25}$
$=\frac{-1}{25}(2-11\text{i})=\frac{1}{25}(-2+11\text{i})$
$\therefore\ \bar{\text{z}}=\frac{1}{25}(-2-11\text{i})=\frac{-2}{25}-\frac{11}{25}\text{i}$
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Question 93 Marks
Solve the system of equations $Re(z^2) = 0, |z| = 2.$
Answer
Given that$, Re(z^2) = 0, |z| = 2$
Let $\text{z}=\text{x}+\text{iy}$ Then $|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$
Given that $\sqrt{\text{x}^2+\text{y}^2}=2$
$\Rightarrow\text{x}^2+\text{y}^2=4$
Also, $\text{z}^2=\text{x}^2+2\text{ixy}-\text{y}^2=(\text{x}^2-\text{y}^2)+2\text{ixy}$
Now, $\text{Re}(\text{z}^2)=0$
$\Rightarrow\text{x}^2-\text{y}^2=0$
Solving $(i)$ and $(ii)$ we get
$\Rightarrow\text{x}^2=\text{y}^2=2$
$\Rightarrow\text{x}=\pm\sqrt{2}$ and $\text{y}=\pm\sqrt{2}$
$\therefore\ \text{z}=\text{x}+\text{iy}=\pm\sqrt{2}\pm\text{i}\sqrt{2}$
Hence$, $we have four complex numbers.
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Question 103 Marks
If $z_1, z_2$ and $z_3, z_4$ are two pairs of conjugate complex numbers$,$ then find $\text{arg}\Big(\frac{\text{z}_1}{\text{z}_4}\Big)+\text{arg}\Big(\frac{\text{z}_2}{\text{z}_3}\Big)$
Answer
It is given that $z_1$ and $z_2$ are conjugate complex numbers.
$\Rightarrow\text{z}_2=\bar{\text{z}}_1$
Also$, z_3$ and $z_4$ are conjugate complex numbers.
$\Rightarrow\text{z}_4=\bar{\text{z}}_3$
Now$, \text{arg}\Big(\frac{\text{z}_1}{\text{z}_4}\Big)+\text{arg}\Big(\frac{\text{z}_2}{\text{z}_3}\Big)=\text{arg}\Big(\frac{\text{z}_1}{\text{z}_4}\Big)\Big(\frac{\text{z}_2}{\text{z}_3}\Big)$
$=\text{arg}\Big(\frac{\text{z}_1}{\text{z}_3}\Big)\Big(\frac{\bar{\text{z}}_1}{\text{z}_3}\Big)=\text{arg}\Big(\frac{\text{z}_1\bar{\text{z}}_1}{\text{z}_3\bar{\text{z}}_3}\Big)$
$=\text{arg}\bigg(\frac{|\text{z}_1|^2}{|\text{z}_3|^2}\bigg)$
$=0$ $\bigg(\because\ \frac{|\text{z}_1|^2}{|\text{z}_3|^2}\text{ is purely real}\bigg)$
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Question 113 Marks
Where does z lie, if $\Big|\frac{\text{z}-5\text{i}}{\text{z}+5\text{i}}\Big|=1$ 
Answer
Given that, $\Big|\frac{\text{z}-5\text{i}}{\text{z}+5\text{i}}\Big|=1$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ \Big|\frac{\text{x}+\text{yi}-5\text{i}}{\text{x}+\text{yi}+5\text{i}}\Big|=1$
$\Rightarrow\Big|\frac{\text{x}+(\text{y}-5)\text{i}}{\text{x}+(\text{y}+5)\text{i}}\Big|=1$
$\Rightarrow|\text{x}+(\text{y}-5)\text{i}|=|\text{x}+(\text{y}+5)\text{i}|$
$\Rightarrow\text{x}^2+(\text{y}-5)^2=\text{x}^2+(\text{y}+5)^2$
$\Rightarrow(\text{y}-5)^2=(\text{y}+5)^2$
$\Rightarrow\text{y}^2+25-10\text{y}=\text{y}^2+25+10\text{y}$
$\Rightarrow20\text{y}=0$
$\Rightarrow\text{y}=0$
Hence, z lies on x-axis i.e., real axis.
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