Complex Numbers and Quadratic EquationsClass 11 Mathematics NCERT Solutions

Given: The complex number expression $(5 i)\left(-\frac{3}{5} i\right)$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We have, $(5 i)\left(-\frac{3}{5} i\right) = 5 \times \left(-\frac{3}{5}\right) \times (i \times i)$ $= -3 \times i^2$ We know that $i^2 = -1$. Substituting this value: $= -3 \times (-1)$ $= 3$ To express this in the form $a+ib$, we can write it as: $3 + 0i$ Here, $a=3$ and $b=0$.

Final Answer: The expression in the form $a+ib$ is $3+0i$.

Given: The complex number expression $i^{9}+i^{19}$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We know the properties of powers of $i$: $i^1 = i$ $i^2 = -1$ $i^3 = -i$ $i^4 = 1$ In general, $i^{4k+r} = i^r$ for any integer $k$.

First, let's simplify $i^9$: $i^9 = i^{4 \times 2 + 1} = i^1 = i$

Next, let's simplify $i^{19}$: $i^{19} = i^{4 \times 4 + 3} = i^3 = -i$

Now, we can find the sum: $i^9 + i^{19} = i + (-i) = 0$

To express this in the form $a+ib$, we can write it as: $0 + 0i$ Here, $a=0$ and $b=0$.

Final Answer: The expression in the form $a+ib$ is $0+0i$.

Given: The complex number expression $i^{-39}$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We have, $i^{-39} = \frac{1}{i^{39}}$ First, let's simplify the denominator $i^{39}$. We know $i^{4k+r} = i^r$. $i^{39} = i^{4 \times 9 + 3} = i^3 = -i$

Now substitute this back into the expression: $i^{-39} = \frac{1}{-i}$ To remove $i$ from the denominator, we multiply the numerator and denominator by $i$: $= \frac{1}{-i} \times \frac{i}{i}$ $= \frac{i}{-i^2}$ Since $i^2 = -1$, we get: $= \frac{i}{-(-1)} = \frac{i}{1} = i$

To express this in the form $a+ib$, we can write it as: $0 + 1i$ Here, $a=0$ and $b=1$.

Final Answer: The expression in the form $a+ib$ is $0+i$.

Given: The complex number expression $3(7+i 7)+i(7+i 7)$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We have, $3(7+i 7)+i(7+i 7)$ First, distribute the terms: $= (3 \times 7 + 3 \times i7) + (i \times 7 + i \times i7)$ $= (21 + 21i) + (7i + 7i^2)$ We know that $i^2 = -1$. Substituting this value: $= 21 + 21i + 7i + 7(-1)$ $= 21 + 21i + 7i - 7$ Now, group the real and imaginary parts: $= (21 - 7) + (21i + 7i)$ $= 14 + 28i$ This is in the form $a+ib$, where $a=14$ and $b=28$.

Final Answer: The expression in the form $a+ib$ is $14+28i$.

Given: The complex number expression $(1-i)-(-1+i 6)$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We have, $(1-i)-(-1+i 6)$ Distribute the negative sign: $= 1 - i - (-1) - (i6)$ $= 1 - i + 1 - 6i$ Now, group the real and imaginary parts: $= (1 + 1) + (-i - 6i)$ $= 2 - 7i$ This is in the form $a+ib$, where $a=2$ and $b=-7$.

Final Answer: The expression in the form $a+ib$ is $2-7i$.

Given: The complex number expression $\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right)$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We have, $\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right)$ First, remove the parentheses: $= \frac{1}{5} + i \frac{2}{5} - 4 - i \frac{5}{2}$ Now, group the real and imaginary parts: $= \left(\frac{1}{5} - 4\right) + i\left(\frac{2}{5} - \frac{5}{2}\right)$ For the real part: $\frac{1}{5} - 4 = \frac{1 - 4 \times 5}{5} = \frac{1 - 20}{5} = -\frac{19}{5}$ For the imaginary part: $\frac{2}{5} - \frac{5}{2} = \frac{2 \times 2 - 5 \times 5}{10} = \frac{4 - 25}{10} = -\frac{21}{10}$ Combining the parts, we get: $= -\frac{19}{5} + i\left(-\frac{21}{10}\right)$ $= -\frac{19}{5} - \frac{21}{10}i$ This is in the form $a+ib$, where $a=-\frac{19}{5}$ and $b=-\frac{21}{10}$.

Final Answer: The expression in the form $a+ib$ is $-\frac{19}{5} - \frac{21}{10}i$.

Given: The complex number expression $\left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)$.

To Find: Express the given complex number in the form $a+i b$.

Solution: First, let's solve the expression inside the square brackets: $\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)$ Group the real and imaginary parts: $= \left(\frac{1}{3} + 4\right) + i\left(\frac{7}{3} + \frac{1}{3}\right)$ $= \left(\frac{1+12}{3}\right) + i\left(\frac{8}{3}\right)$ $= \frac{13}{3} + i\frac{8}{3}$ Now, subtract the third complex number from this result: $\left(\frac{13}{3} + i\frac{8}{3}\right) - \left(-\frac{4}{3}+i\right)$ $= \frac{13}{3} + i\frac{8}{3} + \frac{4}{3} - i$ Group the real and imaginary parts again: $= \left(\frac{13}{3} + \frac{4}{3}\right) + i\left(\frac{8}{3} - 1\right)$ $= \left(\frac{17}{3}\right) + i\left(\frac{8-3}{3}\right)$ $= \frac{17}{3} + i\frac{5}{3}$ This is in the form $a+ib$, where $a=\frac{17}{3}$ and $b=\frac{5}{3}$.

Final Answer: The expression in the form $a+ib$ is $\frac{17}{3} + \frac{5}{3}i$.

Given: The complex number expression $(1-i)^{4}$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We can write $(1-i)^4$ as $((1-i)^2)^2$. First, let's compute $(1-i)^2$: $(1-i)^2 = 1^2 - 2(1)(i) + i^2$ $= 1 - 2i + (-1)$ $= 1 - 2i - 1$ $= -2i$ Now, we need to compute $((1-i)^2)^2$, which is $(-2i)^2$: $(-2i)^2 = (-2)^2 \times (i)^2$ $= 4 \times i^2$ $= 4 \times (-1)$ $= -4$ To express this in the form $a+ib$, we can write it as: $-4 + 0i$ Here, $a=-4$ and $b=0$.

Final Answer: The expression in the form $a+ib$ is $-4+0i$.

Given: The complex number expression $\left(\frac{1}{3}+3 i\right)^{3}$.

To Find: Express the given complex number in the form $a+i b$.

Formula: We use the identity $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$.

Solution: Let $x = \frac{1}{3}$ and $y = 3i$. $\left(\frac{1}{3}+3 i\right)^{3} = \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2(3i) + 3\left(\frac{1}{3}\right)(3i)^2 + (3i)^3$ $= \frac{1}{27} + 3\left(\frac{1}{9}\right)(3i) + 3\left(\frac{1}{3}\right)(9i^2) + (27i^3)$ $= \frac{1}{27} + \frac{1}{3}(3i) + 1(9i^2) + 27i^3$ $= \frac{1}{27} + i + 9i^2 + 27i^3$ We know that $i^2 = -1$ and $i^3 = -i$. Substituting these values: $= \frac{1}{27} + i + 9(-1) + 27(-i)$ $= \frac{1}{27} + i - 9 - 27i$ Now, group the real and imaginary parts: $= \left(\frac{1}{27} - 9\right) + (i - 27i)$ $= \left(\frac{1 - 9 \times 27}{27}\right) - 26i$ $= \left(\frac{1 - 243}{27}\right) - 26i$ $= -\frac{242}{27} - 26i$ This is in the form $a+ib$, where $a=-\frac{242}{27}$ and $b=-26$.

Final Answer: The expression in the form $a+ib$ is $-\frac{242}{27} - 26i$.

Given: The complex number expression $\left(-2-\frac{1}{3} i\right)^{3}$.

To Find: Express the given complex number in the form $a+i b$.

Solution: We can factor out $-1$ from the expression: $\left(-2-\frac{1}{3} i\right)^{3} = \left[-1\left(2+\frac{1}{3} i\right)\right]^{3} = (-1)^3 \left(2+\frac{1}{3} i\right)^{3} = -\left(2+\frac{1}{3} i\right)^{3}$ Now we use the identity $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ with $x=2$ and $y=\frac{1}{3}i$. $\left(2+\frac{1}{3} i\right)^{3} = (2)^3 + 3(2)^2\left(\frac{1}{3}i\right) + 3(2)\left(\frac{1}{3}i\right)^2 + \left(\frac{1}{3}i\right)^3$ $= 8 + 3(4)\left(\frac{1}{3}i\right) + 6\left(\frac{1}{9}i^2\right) + \frac{1}{27}i^3$ $= 8 + \frac{12}{3}i + \frac{6}{9}i^2 + \frac{1}{27}i^3$ $= 8 + 4i + \frac{2}{3}i^2 + \frac{1}{27}i^3$ We know that $i^2 = -1$ and $i^3 = -i$. Substituting these values: $= 8 + 4i + \frac{2}{3}(-1) + \frac{1}{27}(-i)$ $= 8 + 4i - \frac{2}{3} - \frac{1}{27}i$ Group the real and imaginary parts: $= \left(8 - \frac{2}{3}\right) + i\left(4 - \frac{1}{27}\right)$ $= \left(\frac{24 - 2}{3}\right) + i\left(\frac{108 - 1}{27}\right)$ $= \frac{22}{3} + i\frac{107}{27}$ Finally, we multiply by the $-1$ from the beginning: $-\left(\frac{22}{3} + i\frac{107}{27}\right) = -\frac{22}{3} - \frac{107}{27}i$ This is in the form $a+ib$, where $a=-\frac{22}{3}$ and $b=-\frac{107}{27}$.

Final Answer: The expression in the form $a+ib$ is $-\frac{22}{3} - \frac{107}{27}i$.

Given: The complex number $z = 4-3i$.

To Find: The multiplicative inverse of $z$.

Formula: The multiplicative inverse of a complex number $z = a+ib$ is given by $z^{-1} = \frac{1}{z} = \frac{\bar{z}}{|z|^2}$, where $\bar{z} = a-ib$ is the conjugate and $|z|^2 = a^2+b^2$.

Solution: For the given complex number $z = 4-3i$, we have $a=4$ and $b=-3$.

Final Answer: The multiplicative inverse of $4-3i$ is $\frac{4}{25} + \frac{3}{25}i$.

Given: The complex number $z = \sqrt{5}+3i$.

To Find: The multiplicative inverse of $z$.

Formula: The multiplicative inverse of a complex number $z = a+ib$ is given by $z^{-1} = \frac{1}{z} = \frac{\bar{z}}{|z|^2}$, where $\bar{z} = a-ib$ is the conjugate and $|z|^2 = a^2+b^2$.

Solution: For the given complex number $z = \sqrt{5}+3i$, we have $a=\sqrt{5}$ and $b=3$.

Final Answer: The multiplicative inverse of $\sqrt{5}+3i$ is $\frac{\sqrt{5}}{14} - \frac{3}{14}i$.

Given: The complex number $z = -i$.

To Find: The multiplicative inverse of $z$.

Solution: We can write $z$ in the form $a+ib$ as $z = 0 - 1i$. So, $a=0$ and $b=-1$.

Method 1: Using the formula Formula: $z^{-1} = \frac{\bar{z}}{|z|^2}$

Method 2: Direct calculation The multiplicative inverse is $\frac{1}{z}$. $z^{-1} = \frac{1}{-i}$ Multiply the numerator and denominator by $i$ to rationalize: $= \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2}$ Since $i^2 = -1$, we get: $= \frac{i}{-(-1)} = \frac{i}{1} = i$

Final Answer: The multiplicative inverse of $-i$ is $i$.

Given: The expression $\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2} i)-(\sqrt{3}-i \sqrt{2})}$.

To Find: Express the given expression in the form $a+i b$.

Solution: Let's simplify the numerator and the denominator separately.

Numerator: $(3+i \sqrt{5})(3-i \sqrt{5})$ This is of the form $(x+y)(x-y) = x^2 - y^2$. Here $x=3$ and $y=i\sqrt{5}$. $= (3)^2 - (i\sqrt{5})^2$ $= 9 - (i^2 \times (\sqrt{5})^2)$ $= 9 - (-1 \times 5)$ $= 9 - (-5) = 9+5 = 14$

Denominator: $(\sqrt{3}+\sqrt{2} i)-(\sqrt{3}-i \sqrt{2})$ $= \sqrt{3} + \sqrt{2}i - \sqrt{3} + i\sqrt{2}$ $= (\sqrt{3} - \sqrt{3}) + (\sqrt{2}i + i\sqrt{2})$ $= 0 + 2\sqrt{2}i = 2\sqrt{2}i$

Now, combine the simplified numerator and denominator: $\frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i}$ To express this in the form $a+ib$, we need to remove $i$ from the denominator. We multiply the numerator and denominator by $i$: $= \frac{7}{\sqrt{2}i} \times \frac{i}{i}$ $= \frac{7i}{\sqrt{2}i^2}$ Since $i^2 = -1$: $= \frac{7i}{\sqrt{2}(-1)} = -\frac{7i}{\sqrt{2}}$ To rationalize the denominator: $= -\frac{7i}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{7\sqrt{2}i}{2}$ In the form $a+ib$, this is: $0 - \frac{7\sqrt{2}}{2}i$

Final Answer: The expression in the form $a+ib$ is $0 - \frac{7\sqrt{2}}{2}i$.