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

To express the given complex number in the form $a\; +\; ib$, we perform the multiplication:

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

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

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

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 $3\; +\; 0i$.

To simplify $i^9\; +\; i^\{19\}$, we first simplify each term using the property that powers of $i$ repeat in a cycle of 4: $i^1\; =\; i$, $i^2\; =\; -1$, $i^3\; =\; -i$, $i^4\; =\; 1$.

For $i^9$, we divide the exponent 9 by 4: $9\; =\; 4\; \backslash times\; 2\; +\; 1$. So, $i^9\; =\; i^\{4\; \backslash times\; 2\; +\; 1\}\; =\; (i^4)^2\; \backslash cdot\; i^1\; =\; 1^2\; \backslash cdot\; i\; =\; i$.

For $i^\{19\}$, we divide the exponent 19 by 4: $19\; =\; 4\; \backslash times\; 4\; +\; 3$. So, $i^\{19\}\; =\; i^\{4\; \backslash times\; 4\; +\; 3\}\; =\; (i^4)^4\; \backslash cdot\; i^3\; =\; 1^4\; \backslash cdot\; (-i)\; =\; -i$.

Now, we add the simplified terms:

$$i^9\; +\; i^\{19\}\; =\; i\; +\; (-i)\; =\; 0$$

The complex number in the form $a\; +\; ib$ is $0\; +\; 0i$.

To simplify $i^\{-39\}$, we use the property of powers of $i$ and negative exponents. We can rewrite the exponent -39 in terms of multiples of 4.

We can write $-39\; =\; -40\; +\; 1$. So, $i^\{-39\}\; =\; i^\{-40\; +\; 1\}\; =\; i^\{-40\}\; \backslash cdot\; i^1$.

Since $i^4\; =\; 1$, we have $i^\{-40\}\; =\; (i^4)^\{-10\}\; =\; 1^\{-10\}\; =\; 1$.

Therefore, $i^\{-39\}\; =\; 1\; \backslash cdot\; i\; =\; i$.

Alternatively, we can write $-39\; =\; -4\; \backslash times\; 9\; -\; 3$. Then $i^\{-39\}\; =\; i^\{-4\; \backslash times\; 9\; -\; 3\}\; =\; (i^4)^\{-9\}\; \backslash cdot\; i^\{-3\}$.

Since $i^4\; =\; 1$, $(i^4)^\{-9\}\; =\; 1^\{-9\}\; =\; 1$.

And $i^\{-3\}\; =\; \backslash frac\{1\}\{i^3\}\; =\; \backslash frac\{1\}\{-i\}$. To simplify $\backslash frac\{1\}\{-i\}$, multiply numerator and denominator by $i$: $\backslash frac\{1\; \backslash times\; i\}\{-i\; \backslash times\; i\}\; =\; \backslash frac\{i\}\{-i^2\}\; =\; \backslash frac\{i\}\{-(-1)\}\; =\; \backslash frac\{i\}\{1\}\; =\; i$.

So, $i^\{-39\}\; =\; 1\; \backslash cdot\; i\; =\; i$.

The complex number in the form $a\; +\; ib$ is $0\; +\; 1i$.

To express the given complex number in the form $a\; +\; ib$, we first distribute the terms:

$$3(7\; +\; i7)\; =\; 3\; \backslash times\; 7\; +\; 3\; \backslash times\; i7\; =\; 21\; +\; 21i$$

$i(7\; +\; i7)\; =\; i\; \backslash times\; 7\; +\; i\; \backslash times\; i7\; =\; 7i\; +\; 7i^2$

Now, substitute $i^2\; =\; -1$ into the second part:

$7i\; +\; 7i^2\; =\; 7i\; +\; 7(-1)\; =\; 7i\; -\; 7$

Now, add the results of the two distributions:

$$(21\; +\; 21i)\; +\; (7i\; -\; 7)\; =\; 21\; +\; 21i\; +\; 7i\; -\; 7$$

Combine the real parts ($21\; -\; 7$) and the imaginary parts ($21i\; +\; 7i$):

$21\; -\; 7\; =\; 14$

$21i\; +\; 7i\; =\; 28i$

So, the complex number in the form $a\; +\; ib$ is $14\; +\; 28i$.

To express the given complex number in the form $a\; +\; ib$, we first remove the parentheses. Remember to distribute the negative sign to each term inside the second parenthesis:

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

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

Now, combine the real parts ($1\; +\; 1$) and the imaginary parts ($-i\; -\; 6i$):

$1\; +\; 1\; =\; 2$

$-i\; -\; 6i\; =\; -7i$

So, the complex number in the form $a\; +\; ib$ is $2\; -\; 7i$.