CBSE Class 11 Maths Exemplar Chapter 5: Complex Numbers and Quadratic Equations NCERT Solutions

We are asked to find the value of the expression $(1-i)^n\; \backslash left(1-\backslash frac\{1\}\{i\}\backslash right)^n$ for a positive integer n.

First, let's simplify the term $\backslash left(1-\backslash frac\{1\}\{i\}\backslash right)$. We know that $\backslash frac\{1\}\{i\}\; =\; \backslash frac\{1\}\{i\}\; \backslash cdot\; \backslash frac\{i\}\{i\}\; =\; \backslash frac\{i\}\{i^2\}\; =\; \backslash frac\{i\}\{-1\}\; =\; -i$.

Substituting this back into the expression, we get:

$$(1-i)^n\; \backslash left(1\; -\; (-i)\backslash right)^n\; =\; (1-i)^n\; (1+i)^n$$

Now, we can combine the bases since the exponent is the same:

$$=\; [(1-i)(1+i)]^n$$

Using the difference of squares formula, $(a-b)(a+b)\; =\; a^2\; -\; b^2$:

$$=\; [1^2\; -\; i^2]^n$$

Since $i^2\; =\; -1$:

$$=\; [1\; -\; (-1)]^n\; =\; [1+1]^n\; =\; 2^n$$

Thus, the value of the given expression is $2^n$.

We need to evaluate the summation $\backslash sum\_\{n=1\}^\{13\}\; (i^n\; +\; i^\{n+1\})$.

We can factor out $i^n$ from the term inside the summation:

$$\backslash sum\_\{n=1\}^\{13\}\; i^n\; (1+i)$$

Since $(1+i)$ is a constant with respect to the summation index $n$, we can take it out:

$$=\; (1+i)\; \backslash sum\_\{n=1\}^\{13\}\; i^n$$

Now, let's evaluate the summation $\backslash sum\_\{n=1\}^\{13\}\; i^n$. The powers of $i$ follow a cycle: $i^1=i,\; i^2=-1,\; i^3=-i,\; i^4=1$. The sum of any four consecutive powers of $i$ is zero, i.e., $i^n\; +\; i^\{n+1\}\; +\; i^\{n+2\}\; +\; i^\{n+3\}\; =\; 0$.

The summation is $i^1\; +\; i^2\; +\; i^3\; +\; \backslash dots\; +\; i^\{13\}$. We can group these terms in sets of four:

$\backslash sum\_\{n=1\}^\{13\}\; i^n\; =\; (i^1\; +\; i^2\; +\; i^3\; +\; i^4)\; +\; (i^5\; +\; i^6\; +\; i^7\; +\; i^8)\; +\; (i^9\; +\; i^\{10\}\; +\; i^\{11\}\; +\; i^\{12\})\; +\; i^\{13\}$

Each group of four sums to zero:

$=\; 0\; +\; 0\; +\; 0\; +\; i^\{13\}$

To find $i^\{13\}$, we divide 13 by 4: $13\; =\; 4\; \backslash times\; 3\; +\; 1$. So, $i^\{13\}\; =\; (i^4)^3\; \backslash cdot\; i^1\; =\; 1^3\; \backslash cdot\; i\; =\; i$.

Therefore, $\backslash sum\_\{n=1\}^\{13\}\; i^n\; =\; i$.

Substituting this back into our expression:

$$=\; (1+i)\; \backslash cdot\; i$$

$$=\; i\; +\; i^2$$

$$=\; i\; -\; 1$$

The value of the summation is $i-1$.

We are given the equation $\backslash left(\backslash frac\{1+i\}\{1-i\}\backslash right)^3\; -\; \backslash left(\backslash frac\{1-i\}\{1+i\}\backslash right)^3\; =\; x+iy$ and need to find the values of $x$ and $y$.

First, let's simplify the complex fractions:

For the first term, $\backslash frac\{1+i\}\{1-i\}$:

$$\backslash frac\{1+i\}\{1-i\}\; =\; \backslash frac\{(1+i)\}\{(1-i)\}\; \backslash cdot\; \backslash frac\{(1+i)\}\{(1+i)\}\; =\; \backslash frac\{(1+i)^2\}\{1^2\; -\; i^2\}\; =\; \backslash frac\{1\; +\; 2i\; +\; i^2\}\{1\; -\; (-1)\}\; =\; \backslash frac\{1\; +\; 2i\; -\; 1\}\{2\}\; =\; \backslash frac\{2i\}\{2\}\; =\; i$$

So, $\backslash left(\backslash frac\{1+i\}\{1-i\}\backslash right)^3\; =\; i^3\; =\; -i$.

For the second term, $\backslash frac\{1-i\}\{1+i\}$:

$$\backslash frac\{1-i\}\{1+i\}\; =\; \backslash frac\{(1-i)\}\{(1+i)\}\; \backslash cdot\; \backslash frac\{(1-i)\}\{(1-i)\}\; =\; \backslash frac\{(1-i)^2\}\{1^2\; -\; i^2\}\; =\; \backslash frac\{1\; -\; 2i\; +\; i^2\}\{1\; -\; (-1)\}\; =\; \backslash frac\{1\; -\; 2i\; -\; 1\}\{2\}\; =\; \backslash frac\{-2i\}\{2\}\; =\; -i$$

So, $\backslash left(\backslash frac\{1-i\}\{1+i\}\backslash right)^3\; =\; (-i)^3\; =\; -(i^3)\; =\; -(-i)\; =\; i$.

Now, substitute these simplified values back into the original equation:

$$(-i)\; -\; (i)\; =\; x+iy$$

$$-2i\; =\; x+iy$$

To find $x$ and $y$, we compare the real and imaginary parts of the equation. The left side, $-2i$, can be written as $0\; +\; (-2)i$.

Comparing the real parts: $x\; =\; 0$.

Comparing the imaginary parts: $y\; =\; -2$.

Therefore, the value of $(x,\; y)$ is $(0,\; -2)$.

We are given the equation $\backslash frac\{(1+i)^2\}\{2\}\; =\; x+iy$ and asked to find the value of $x+y$.

First, let's simplify the left side of the equation:

$$\backslash frac\{(1+i)^2\}\{2\}\; =\; \backslash frac\{1^2\; +\; 2(1)(i)\; +\; i^2\}\{2\}$$

Since $i^2\; =\; -1$:

$$=\; \backslash frac\{1\; +\; 2i\; -\; 1\}\{2\}\; =\; \backslash frac\{2i\}\{2\}\; =\; i$$

So, the equation becomes:

$$i\; =\; x+iy$$

To find $x$ and $y$, we compare the real and imaginary parts. The complex number $i$ can be written as $0\; +\; 1i$.

Comparing the real parts: $x\; =\; 0$.

Comparing the imaginary parts: $y\; =\; 1$.

We are asked to find the value of $x+y$.

$$x+y\; =\; 0\; +\; 1\; =\; 1$$

The value of $x+y$ is 1.