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

To express the complex number $(5i)\left(-\frac{3}{5}i\right)$ in the form $a + ib$, we perform the multiplication:

(5i)\left(-\frac{3}{5}i\right) = 5 \times \left(-\frac{3}{5}\right) \times i \times i

First, multiply the numerical coefficients: $5 \times \left(-\frac{3}{5}\right) = -3$.

Next, multiply the imaginary units: $i \times i = i^2$.

So, the expression becomes $-3i^2$.

We know that $i^2 = -1$. Substituting this value:

-3i^2 = -3(-1) = 3 Therefore, the complex number in the form $a + ib$ is $3 + 0i$. Here, $a=3$ and $b=0$.

To express $i^9 + i^{19}$ in the form $a + ib$, we first simplify the powers of $i$. We use the property that $i^4 = 1$.

For $i^9$: Divide the exponent 9 by 4. $9 = 4 \times 2 + 1$. So, $i^9 = i^{4 \times 2 + 1} = (i^4)^2 \cdot i^1$. Since $i^4 = 1$, this becomes $1^2 \cdot i = i$.

For $i^{19}$: Divide the exponent 19 by 4. $19 = 4 \times 4 + 3$. So, $i^{19} = i^{4 \times 4 + 3} = (i^4)^4 \cdot i^3$. Since $i^4 = 1$ and $i^3 = -i$, this becomes $1^4 \cdot (-i) = -i$.

Now, add the simplified terms:

i^9 + i^{19} = i + (-i) = 0 Therefore, the complex number in the form $a + ib$ is $0 + 0i$. Here, $a=0$ and $b=0$.

To express $i^{-39}$ in the form $a + ib$, we first simplify the negative power of $i$. We can write $i^{-39}$ as $\frac{1}{i^{39}}$.

Now, let's simplify $i^{39}$. We use the property $i^4 = 1$. Divide the exponent 39 by 4: $39 = 4 \times 9 + 3$.

So, $i^{39} = i^{4 \times 9 + 3} = (i^4)^9 \cdot i^3$. Since $i^4 = 1$ and $i^3 = -i$, we have:

$$i^\{39\}\; =\; (1)^9\; \backslash cdot\; (-i)\; =\; 1\; \backslash cdot\; (-i)\; =\; -i$$

Now, substitute this back into the expression for $i^{-39}$:

$$i^\{-39\}\; =\; \backslash frac\{1\}\{i^\{39\}\}\; =\; \backslash frac\{1\}\{-i\}$$

To express this in the form $a + ib$, we multiply the numerator and denominator by $i$ (or by $-i$ to rationalize):

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

Since $i^2 = -1$, we get:

$$\backslash frac\{i\}\{-(-1)\}\; =\; \backslash frac\{i\}\{1\}\; =\; i$$

Therefore, the complex number in the form $a + ib$ is $0 + 1i$. Here, $a=0$ and $b=1$.

To express $3(7 + i7) + i(7 + i7)$ in the form $a + ib$, we first distribute the terms:

Multiply 3 by each term inside the first parenthesis: $3 \times 7 = 21$ and $3 \times i7 = 21i$. So, $3(7 + i7) = 21 + 21i$.

Multiply $i$ by each term inside the second parenthesis: $i \times 7 = 7i$ and $i \times i7 = 7i^2$. So, $i(7 + i7) = 7i + 7i^2$.

Now, add the results:

$$3(7\; +\; i7)\; +\; i(7\; +\; i7)\; =\; (21\; +\; 21i)\; +\; (7i\; +\; 7i^2)$$

Combine the terms: $21 + 21i + 7i + 7i^2$.

Group the real and imaginary terms: $21 + (21i + 7i) + 7i^2$.

This simplifies to $21 + 28i + 7i^2$.

Substitute $i^2 = -1$: $21 + 28i + 7(-1)$.

This becomes $21 + 28i - 7$.

Combine the real parts: $(21 - 7) + 28i$.

$$=\; 14\; +\; 28i$$

Therefore, the complex number in the form $a + ib$ is $14 + 28i$. Here, $a=14$ and $b=28$.

To express $(1 - i) - (-1 + i6)$ in the form $a + ib$, we first remove the parentheses. Remember to distribute the negative sign to both terms inside the second parenthesis.

$$(1\; -\; i)\; -\; (-1\; +\; i6)\; =\; 1\; -\; i\; -\; (-1)\; -\; (i6)$$

This simplifies to:

$$1\; -\; i\; +\; 1\; -\; 6i$$

Now, group the real terms and the imaginary terms:

Real terms: $1 + 1 = 2$.

Imaginary terms: $-i - 6i = (-1 - 6)i = -7i$.

Combine the real and imaginary parts:

$$2\; -\; 7i$$

Therefore, the complex number in the form $a + ib$ is $2 - 7i$. Here, $a=2$ and $b=-7$.