Math Assignment Class XI Ch-04

Math Assignment Class XI Ch - 04

Extra questions of chapter 04 COMPLEX NUMBERS class 11 with answers and hints to the difficult questions, strictly according to the CBSE & DAV Board syllabus.

ASSIGNMENT ON COMPLEX NUMBERS (XI)

Strictly according to the CBSE and DAV Board

Express the following complex number in a + ib form

Question1: i⁹+ i¹⁰+ i¹¹+ i¹².

Answer: 0 + 0i

Question 2: (7 - 2i) - (4 + i) + (- 3 + 5i)

Answer: 0 + 2i

Question 3: (2 + 3i) (2 - 3i)(1 + i²)

Answer: 0

Question 4:

Evaluate: $\mathbf{\frac{\left ( \sqrt{2}+i\sqrt{3} \right )+\left ( \sqrt{2}-i\sqrt{3} \right )}{\left ( \sqrt{3}+i\sqrt{2} \right )+\left ( \sqrt{3}-i\sqrt{2} \right )}}$

Answer: $\mathbf{\sqrt{6}/3+0i}$

Question 5:

Evaluate: $\mathbf{\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}}$

Answer: -1 + 0i

Question 6: If 3 + yi - 2i = x - i. Find y.

Answer: x = 3, y = 1

Question 7: Find the modulus of (1 - i)¹⁰.

Answer: 32

Question 8: Find the conjugate of $\mathbf{\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}}$

Answer: $\mathbf{\frac{63}{25}+\frac{16}{25}i}$

Question 9:

$\mathbf{If\: \: \left ( \frac{1+i}{1-i} \right )^{3}-\left ( \frac{1-i}{1+i} \right )^{3}=x+iy,\: \: find\: \: x,\: y}$

Answer: x = 0, y = -2

Question 10:

Evaluate: 2x³ + 2x² - 7x + 72, when $\mathbf{ x=\frac{3-5i}{2}}$

Answer: 4

Solution Hint $\mathbf{ x=\frac{3-5i}{2}}$

2x = 3 - 5i

2x - 3 = - 5i

Squaring on both side and arranging all terms on the one side of equal we get

2x² - 6x + 17 = 0 ...... (i)

Now divide the given cubic equation by eqn. (i) we get quotient = x+4 and remainder = 4

2x³ + 2x² - 7x + 72 = (2x² - 6x + 17)(x + 4) + 4

= 0 × (x + 4) + 4

= 4

Question 11: Evaluate: x⁴ + 4x³ + 6x² + 4x + 9, where x = -1+ i√ 2

Answer: 12

Question 12:

If z = 2 - 3i, then show that z² - 4z + 13 = 0. Hence, find the value of 4z³- 3z²+ 2z + 170.

Answer: 5 - 6i

Question 13:

Find the real values of x and y if $\mathbf{\frac{x+iy}{1+i}}$ is the conjugate of 5 - i

Answer: x = 4, y = 6

Question 14: Evaluate: $\mathbf{\sqrt{-16}+\sqrt{-25}-\sqrt{-36}+\sqrt{-625}}$

Answer: 28i

Question 15: Find two real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.

Question 16: Write the conjugate of complex number in the form a + ib. $\mathbf{\frac{1}{1+i}}$

Answer: $\mathbf{\frac{1+i}{2}}$

Question 17: If x + iy= $\mathbf{\frac{(a^{2}+1)^{2}}{2a-i}}$ , then find the value of x² + y².

Answer: x² + y² = $\mathbf{\frac{(a^{2}+1)^{4}}{4a^{2}+1}}$

Question 18: If (x - iy)^1/3 = a +ib, x, y, a ∊ R, then show that $\mathbf{\frac{x}{a}+\frac{y}{b}=-2(a^{2}+b^{2})}$

Question 19: If (x + iy)³ = u + iv, then show that $\mathbf{\frac{u}{x}+\frac{v}{y}=4(x^{2}-y^{2})i=\sqrt{-1}}$

Question 20: If $\mathbf{2p+iq=\frac{a+ib}{a-ib}$ , then show that 4p² + q² =1

Question 21: Evaluate : $\mathbf{\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)}$ , n ∈ N, where $\mathbf{i=\sqrt{-1}}$

Answer: i

Solution Hint:

$\mathbf{\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)=\sum_{n=1}^{13}i^{n}(1+i)\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)}$

= i¹(1 + i) + i²(1 + i) + i³(1 + i) + i⁴(1 + i) + ……… 13 terms

= i

Question 22: If (1 + i)(1 + 2i)(1 + 3i)......(1 + ni) = a + ib . Prove that 2.5.10.....(1+ n² ) = a² + b².

****************************************

FOLLOWING QUESTIONS ARE DELETED FROM CBSE SYLLABUS

Question 19:

Express the complex numbers in standard form and hence convert it in polar form.

$Z=\frac{1-i}{1+i}-\frac{1+i}{1-i}$

Answer: Z = 0 - 2i

Polar form is : $2\left [ cos\left ( -\frac{\pi }{2} \right )+isin\left ( -\frac{\pi }{2} \right ) \right ]$

Question 20:

Convert the complex number 2 - 2i into polar form. Also write its argument.

Question 21: Find the square root of the complex number 4 - 4√3i

Question 22: Find the square root of Z = $\frac{-22+19i}{2+i}$

Question 23: Convert the complex number $3\left [ cos\frac{5\pi }{3}-isin\frac{\pi }{6} \right ]$ into polar form.

Question 24: Find principal argument of the complex number $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$

Answer: $\theta =\pi /4$

THANKS FOR YPUR VISIT

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