05 ਕੰਪਲੈਕਸ ਨੰਬਰ ਅਤੇ ਕੁਆਡ੍ਰੈਟਿਕ ਸਮੀਕਰਨ
ਅਭਿਆਸ 01
Question:
Find the multiplicative inverse of the complex number −i
Answer:
Answer: Step 1: The multiplicative inverse of any number is the reciprocal of the number.
Step 2: The reciprocal of -i is 1/-i.
Step 3: Therefore, the multiplicative inverse of -i is 1/-i.
Question:
Express the given complex number in the form a+ib:(1−i)4
Answer:
(1-i)^4 = (1-i)(1-i)(1-i)(1-i) = (1-i)(1-i)(1-i-i) = (1-i)(1-2i) = (1-2i-i^2) = (1-2i-1) = 1-2i-1+i^2 = 1-2i-1+(-1) = 1-2i-2 = 1-2(i+1) = 1-2i-2i = 1-4i = a+ib = 1-4i
Therefore, the complex number in the form a+ib is 1-4i.
Question:
Find the multiplicative inverse of the complex number √5+3i.
Answer:
Answer: Step 1: Find the conjugate of the complex number √5+3i. Conjugate = √5-3i
Step 2: Find the modulus of the complex number √5+3i. Modulus = |√5+3i| = √(5^2 + 3^2) = √34
Step 3: Find the multiplicative inverse of the complex number √5+3i. Multiplicative inverse = (√5-3i)/√34
Question:
(5i)(−3/5i)
Answer:
-
First, combine the two terms by adding them: 5i + (-3/5i)
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Simplify the terms by multiplying the numerator and denominator of the second term by the complex conjugate of the denominator of the second term: 5i + (-3/(5i)(-5i)) = 5i + (15/25)
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Simplify the terms by combining the coefficients of the terms: 5i + 15/25 = (5i + 15)/25
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Simplify the terms by multiplying the numerator and denominator by the reciprocal of the denominator: (5i + 15)/25 * 25/25 = (5i + 15)/25 * 1 = (5i + 15)
Question:
Solve : i9+i19
Answer:
Step 1: Rewrite the expression using the power of a power rule.
i^9 + i^19 = i^(9+19)
Step 2: Calculate the value of i^(9+19).
i^(9+19) = i^28
Question:
Express the given complex number 3(7+i7)+i(7+i7) in the form a+ib,
Answer:
Given: 3(7+i7)+i(7+i7)
Step 1: 3(7+i7) = 21 + 3i7
Step 2: i(7+i7) = -7 + i7
Step 3: 21 + 3i7 + (-7 + i7)
Step 4: 21 - 7 + 4i7
Answer: 21 - 7 + 4i7 = 14 + 4i7
Question:
Express the given complex number in the form a+ib: (−2−1/3i)3
Answer:
Given complex number, z = (-2 - i/3)
Step 1: Calculate z^3
z^3 = (-2 - i/3)^3
Step 2: Expand the expression
(-2 - i/3)^3 = -8 - 8i/3 - 2i^2/3 - (i/3)^3
Step 3: Simplify the expression
-8 - 8i/3 - 2i^2/3 - (i/3)^3 = -8 - 8i/3 - 2i^2/3 - i/27
Step 4: Express in the form a+ib
-8 - 8i/3 - 2i^2/3 - i/27 = -8 - 8i/3 - 2i^2/3 - i/27 = -8 - 8i/3 - 2i - i/9 = -8 - 10i - i/9
Therefore, the given complex number in the form a+ib is: -8 - 10i - i/9
Question:
Express the following expression in the form of a+ib ; (3+i√5)(3−i√5)/(√3+√2i)−(√3−i√2)
Answer:
Given, (3+i√5)(3−i√5)/(√3+√2i)−(√3−i√2)
Step 1: Multiply the numerator and denominator by the conjugate of the denominator.
(3+i√5)(3−i√5)(√3-√2i)/((√3+√2i)(√3-√2i))-((√3−i√2)(√3-√2i))
Step 2: Simplify the numerator and denominator.
(3+i√5)(3−i√5)(3+√2i)/(9+2i√6)-(3+√2i)(3+√2i)
Step 3: Simplify the numerator and denominator further.
(9-5√5)/(9+2i√6)-9
Step 4: Simplify the expression further.
-14-2i√6/9+2i√6
Step 5: Express the expression in the form of a+ib.
-14/9+2i√6/9
Question:
Solve: [(1/3+i7/3)+(4+i1/3)]−(−4/3+i)
Answer:
Step 1: Combine like terms on the left side of the equation: [(1/3+i7/3)+(4+i1/3)]−(−4/3+i)
Step 2: [(5/3+i8/3)]−(−4/3+i)
Step 3: [5/3+i8/3 + 4/3-i]
Step 4: 9/3+i7/3
Question:
Express (31+3i)3 in the form a+ib. A 343/34+23i B −343/34−23i C 242/27+26i D −242/27−26i
Answer:
Answer: C 242/27+26i
Question:
Find the modulus and amplitude of 4+3i.
Answer:
Answer: Modulus: 5 Amplitude: 5
Question:
Express (1−i)−(1+i6) as a+ib
Answer:
(1−i)−(1+i6) = (1−i) − 1 - i6 = -i - 1 - i6 = -i - 1 - (i/6) = -i - 1 - (1/6)i = -i - 1 + (-1/6)i = -i - 1 + (1/6)i = (-1 - 1 + (1/6))i = (-2 + (1/6))i = (-2 + (1/6))(1 + i) = (-2 + (1/6)) + (-2 + (1/6))i = (-2 + (1/6)) + i(-2 + (1/6)) = a + ib = (-2 + (1/6)) + i(-2 + (1/6)) = (-2 + (1/6)) + i(-2 + (1/6))
Question:
Solve the problem:- (1/5+i2/5)−(4+i5/2)
Answer:
Given, (1/5 + i2/5) - (4+i5/2)
Step 1: Rewrite the given expression as,
(1/5 + i2/5) - (4+ i5/2) = (1/5 - 4) + (i2/5 - i5/2)
Step 2: Simplify the expression,
(1/5 - 4) + (i2/5 - i5/2) = -3.8 + (-3.5i)
Hence, the solution is -3.8 + (-3.5i).
Question:
Evaluate i−39.
Answer:
i^−39 = 1/i^39
= 1/(i^2)^19
= 1/i^38
= 1/i^2 * 1/i^36
= 1/i^2 * (1/i^18)^2
= 1/i^2 * (1/i^2)^18
= (1/i^2)^19
ਜੇਈਈ ਅਧਿਐਨ ਸਮੱਗਰੀ (ਗਣਿਤ)
01 ਸੈੱਟ
02 ਸਬੰਧ ਅਤੇ ਕਾਰਜ
03 ਤ੍ਰਿਕੋਣਮਿਤੀਕ ਫੰਕਸ਼ਨ
04 ਗਣਿਤਿਕ ਇੰਡਕਸ਼ਨ ਦਾ ਸਿਧਾਂਤ
05 ਕੰਪਲੈਕਸ ਨੰਬਰ ਅਤੇ ਕੁਆਡ੍ਰੈਟਿਕ ਸਮੀਕਰਨ
06 ਰੇਖਿਕ ਅਸਮਾਨਤਾਵਾਂ
07 ਪਰਮਿਊਟੇਸ਼ਨ ਅਤੇ ਕੰਬੀਨੇਸ਼ਨ
08 ਬਾਇਨੋਮਿਅਲ ਥਿਊਰਮ
09 ਕ੍ਰਮ ਅਤੇ ਲੜੀ
10 ਸਿੱਧੀਆਂ ਲਾਈਨਾਂ ਦੀ ਕਸਰਤ
10 ਸਿੱਧੀਆਂ ਰੇਖਾਵਾਂ ਫੁਟਕਲ
11 ਕੋਨਿਕ ਸੈਕਸ਼ਨ
12 ਤਿੰਨ ਅਯਾਮੀ ਜਿਓਮੈਟਰੀ ਦੀ ਜਾਣ-ਪਛਾਣ
13 ਸੀਮਾਵਾਂ ਅਤੇ ਡੈਰੀਵੇਟਿਵਜ਼
14 ਗਣਿਤਿਕ ਤਰਕ
15 ਅੰਕੜੇ
16 ਸੰਭਾਵਨਾ