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MATHS 2 Marks Questions

Complex Numbers and Quadratic Equations · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 6 questions.

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Question 12 Marks
Find z if |z| = 4 and $\arg(\text{z})=\frac{5\pi}{6}$
Answer
Let $\text{z}=|\text{z}|(cos\theta+\text{i}\sin\theta),$ where $\theta=\arg(\text{z})$
Given that, |z| = 4 and$\arg(\text{z})=\frac{5\pi}{6}$
$\Rightarrow\text{z}=4\Big[\cos\frac{5\pi}{6}+\text{i}\sin\frac{5\pi}{6}\Big]$ (z lies in II quadrant)
$\Rightarrow\text{z}=4\Big[-\frac{\sqrt{3}}{2}+\text{i}\frac{1}{2}\Big]$
$\Rightarrow\text{z}=-2\sqrt{3}+2\text{i}$ 
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Question 22 Marks
If |z1| = |z2|, is it necessary that z1 = z2?
Answer
If |z1| = |z2| then z1 and z2 are at the same distance from origin.
But if arg(z1) ≠ arg(z2), then z1 and z2 are different.
So, if (|z1| = |z2|), then it is not necessary that z1 = z2
Consider z1 = 3 + 4i and z2 = 4 + 3i.
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Question 32 Marks
If $\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{10}=\text{a}+\text{ib},$ then find (a, b).
Answer
$\text{a}+\text{ib}=\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{10}$
$=\bigg[\frac{(1-\text{i})}{(1+\text{i})}\cdot\frac{(1-\text{i})}{(1-\text{i})}\bigg]^{100}=\bigg[\frac{(1-\text{i})^2}{1-\text{i}^2}\bigg]^{100}$
$=\Big(\frac{1-2\text{i}+\text{i}^2}{1+1}\Big)^{100}=\Big(\frac{-2\text{i}}{2}\Big)^{100}$
$=(\text{i}^{4})^{25}=1$
$\therefore\ (\text{a},\text{b})=(1,0)$
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Question 42 Marks
If $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^3-\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^3=\text{x}+\text{iy},$ then find (x, y).
Answer
$\text{x}+\text{iy}=\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^3-\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^3$
$=\bigg(\frac{(1+\text{i})^2}{1-\text{i}^2}\bigg)-\bigg(\frac{(1-\text{i})^2}{1-\text{i}^2}\bigg)^3$
$=\Big(\frac{1+2\text{i}+\text{i}^2}{1+1}\Big)^3-\Big(\frac{1-2\text{i}+\text{i}^2}{1+1}\Big)^3$
$=\Big(\frac{2\text{i}}{2}\Big)^3-\Big(\frac{-2\text{i}}{2}\Big)^3$
$=\text{i}^3-(-\text{i})^3$
$=2\text{i}^3=0-2\text{i}$
$\text{x}=0\text{ and }\text{y}=-2$
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Question 52 Marks
If $\text{z}=\text{x}+\text{iy},$ then show that $\text{z}\bar{\text{z}}+2(\text{z}+\bar{\text{z}})+\text{b}=0,$ where $\text{b}\in\text{R},$ represents a circle.
Answer
Given that $\text{z}=\text{x}+\text{iy}$
$\Rightarrow\bar{\text{z}}=\text{x}-\text{iy}$
Now,$\text{z}\bar{\text{z}}+2(\text{z}+\bar{\text{z}})+\text{b}=0$
$\Rightarrow(\text{x}+\text{iy})(\text{x}-\text{iy})+2(\text{x}+\text{iy}+\text{x}-\text{iy})+\text{b}=0$
$\Rightarrow\text{x}^2+\text{y}^2+4\text{x}+\text{b}=0$ this is the equation of a circle.
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Question 62 Marks
Find principal argument of $(1+\text{i}\sqrt{3})^2$
Answer
We have,
$\text{z}=(1+\text{i}\sqrt{3})^2$
$\Rightarrow\text{z}=1-3+2\text{i}\sqrt{3}$
$\Rightarrow\text{z}=-2+\text{i}(2\sqrt{3})$
So, z lies in second quadrant.
$\Rightarrow\arg(\text{z})=\pi-\tan^{-1}\Big|\frac{2\sqrt{3}}{-2}\Big|$
$\Rightarrow\arg(\text{z})=\pi-\tan^{-1}\sqrt{3}$
$\Rightarrow\arg(\text{z})=\pi-\frac{\pi}{3}$
$\Rightarrow\arg(\text{z})=\frac{2\pi}{3}$
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