Question:

Find the real numbers x and y , if (x−iy)(3+5i) is the conjugate of −6−24i.

Answer:

1. First, we need to find the conjugate of −6−24i. This is just the complex number with the same real part and the imaginary part with the opposite sign. So, the conjugate of −6−24i is −6+24i.

2. Next, we need to find the complex number (x−iy)(3+5i) that is equal to the conjugate of −6−24i. We can do this by expanding the expression and equating the real and imaginary parts:

(x−iy)(3+5i) = (3x−5y)+(5x+3y)i

3. Now, we can set up two equations with two unknowns, x and y, to solve for these values:

3x−5y = −6 5x+3y = 24

4. We can solve this system of equations using any method. For example, we can use substitution to solve for x:

3x−5y = −6 5x+3y = 24

5y = 30 x = −10

5. Finally, we can substitute x = −10 into the other equation to solve for y:

5x+3y = 24 5(−10)+3y = 24 −50+3y = 24 3y = 74 y = 24.67

Question:

Solve the equation : 27x^2−10x+1=0

Answer:

1. Divide by 27: x^2- (10/27)x + (1/27) = 0

2. Use the quadratic formula: x = (10 ± √(100 - 108))/54

3. Simplify: x = (10 ± 2√3)/54

4. Solve: x = (10 + 2√3)/54 or x = (10 - 2√3)/54

Question:

For any two complex numbers z1​ and z2​, prove that Re(z1​z2​)=Rez1​Rez2​−Imz1​Imz2

Answer:

Given: z1 = a + ib, z2 = c + id

1. Re(z1z2) = Re(ac + ibc + iad - bd)

2. Re(z1z2) = ac - bd

3. Re(z1z2) = Re(z1)Re(z2) - Im(z1)Im(z2)

4. Re(z1z2) = (a + ib)(c + id)

5. Re(z1z2) = ac + bcd + iad - bd

6. Re(z1z2) = ac - bd

7. Re(z1z2) = Re(z1)Re(z2) - Im(z1)Im(z2)

Therefore, it has been proven that Re(z1z2) = Re(z1)Re(z2) - Im(z1)Im(z2).

Question:

Evalaute : [i^18+(1​/i)^25]^3

Answer:

Answer: Step 1: [i^18+(1/i)^25]^3 = (i^18+i^-25)^3

Step 2: Using the laws of exponents, (i^18+i^-25)^3 = i^(18+3) + i^(-25+3)

Step 3: i^(18+3) + i^(-25+3) = i^21 + i^-22

Question:

If a+ib=(x+i)^2​/2x^2+1, prove that a^2+b^2=(x^2+1)^2​/(2x^2+1)^2.

Answer:

Given: a+ib = (x+i)^2/2x^2+1

Step 1: Multiply both sides of the equation by 2x^2+1.

a+ib(2x^2+1)= (x+i)^2

Step 2: Square both sides of the equation.

(a+ib)^2(2x^2+1)^2 = (x+i)^4

Step 3: Expand the left side of the equation.

a^2(2x^2+1)^2 + 2ab(2x^2+1) + b^2(2x^2+1)^2 = (x+i)^4

Step 4: Expand the right side of the equation.

a^2(2x^2+1)^2 + 2ab(2x^2+1) + b^2(2x^2+1)^2 = x^4 + 2ix^3 + 2x^2 + i^4

Step 5: Equate the coefficients of corresponding terms on both sides of the equation.

a^2(2x^2+1)^2 = x^4 2ab(2x^2+1) = 2ix^3 b^2(2x^2+1)^2 = 2x^2 + i^4

Step 6: Solve for a^2 and b^2.

a^2 = x^4/(2x^2+1)^2 b^2 = (2x^2+1)^2/(2x^2+1)^2

Step 7: Substitute the values of a^2 and b^2 in the equation a^2+b^2.

a^2+b^2 = x^4/(2x^2+1)^2 + (2x^2+1)^2/(2x^2+1)^2

Step 8: Simplify.

a^2+b^2 = (x^4 + (2x^2+1)^2)/(2x^2+1)^2

Step 9: Factorize both numerator and denominator.

a^2+b^2 = (x^2+1)^2/(2x^2+1)^2

Hence, proved.

Question:

If z1​=2−i,z2​=1+i, find ∣​z1​+z2​+1/z1​−z2​+1​∣

Answer:

Step 1: Find z1 + z2 + 1/z1 - z2 + 1 z1 + z2 + 1/z1 - z2 + 1 = (2 - i) + (1 + i) + (1/(2 - i)) - (1 + i) + 1

Step 2: Simplify (2 - i) + (1 + i) + (1/(2 - i)) - (1 + i) + 1 = 2 + 2 - i - i + (1/(2 - i)) + 1 = 4 + (1/(2 - i)) + 1

Step 3: Find the absolute value |4 + (1/(2 - i)) + 1| = |4 + (1/(2 - i)) + 1| = |5 + (1/(2 - i))| = 5 + (1/(2 - i))

Question:

Solve the equation : x^2−2x+3​/2=0

Answer:

Step 1: Rewrite the equation as:

(x^2 - 2x + 3)/2 = 0

Step 2: Multiply both sides by 2:

x^2 - 2x + 3 = 0

Step 3: Use the quadratic formula to solve for x:

x = [2 ± √(4 - 12)]/2

Step 4: Simplify:

x = 1 ± √(-3)/2

Question:

Convert the following in the polar form: (i) 1+7i​/(2−i)^2 (ii) 1+3i/1−2i

Answer:

(i) 1+7i/(2−i)^2 = (1+7i)/((2-i)(2+i)) = (1+7i)/(4+2i-2i-1) = (1+7i)/(3+i) = (1/3+7i/3)/(1+i) = (1/3+7i/3)(1-i)/(1+i)(1-i) = (1/3+7i/3)(1-i)/(1+i^2) = (1/3+7i/3)(1-i)/2 = (1/6+7i/6)/(1+i)

Polar form: (1/6+7i/6)∠tan^-1(1/1)

(ii) 1+3i/1−2i = (1+3i)/(1-2i) = (1+3i)(1+2i)/(1-2i)(1+2i) = (1+3i+2i+6)/(1+4) = (1+5i+6)/5 = (1/5+5i/5)/(1+2i/5)

Polar form: (1/5+5i/5)∠tan^-1(2/5)

Question:

Let z1​=2−i,z2​=−2+i. Find (i) Re(z1​z2​​/z1​ˉ​) (ii) Im(1​/z1​z1​ˉ​)

Answer:

(i) Re(z1z2/z1ˉ) = Re((2−i)(−2+i)/(2+i)) = Re((−4−4i)/(2+i)) = Re((−4−4i)(2−i)/(2+i)(2−i)) = Re((−8+4i)/5) = −8/5

(ii) Im(1/z1z1ˉ) = Im(1/(2−i)(2+i)) = Im((2+i)/5) = 4/5

Question:

If (a+ib)(c+id)(e+if)(g+ih)=A+iB, then show that (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2.

Answer:

Step 1: Expand (a+ib)(c+id)(e+if)(g+ih) Step 2: (ac+adib+bcei+bdfi^2)(eg+ehib+fghi+fhi^2) Step 3: aceg+acehib+adbcei+adbdfi^2+bcegfi+bcehi^2+bdfegi+bdfhi^3 Step 4: aceg+acehib+adbcei+adbdfi^2+bcegfi+bcehi^2+bdfegi+bdfhi^3=A+iB Step 5: Multiply both sides by the conjugate of (a+ib)(c+id)(e+if)(g+ih). Step 6: (aceg+acehib+adbcei+adbdfi^2+bcegfi+bcehi^2+bdfegi+bdfhi^3)(aceg-acehib+adbcei-adbdfi^2+bcegfi-bcehi^2+bdfegi-bdfhi^3) Step 7: (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=(aceg+acehib+adbcei+adbdfi^2+bcegfi+bcehi^2+bdfegi+bdfhi^3)(aceg-acehib+adbcei-adbdfi^2+bcegfi-bcehi^2+bdfegi-bdfhi^3) Step 8: (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2 Step 9: Hence, it is proved that (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2.

Question:

Solve the equation : 3x^2−4x+20​/3=0

Answer:

Step 1: Multiply both sides by 3. 3x^2−4x+20​/3 = 0 * 3 3x^2−4x+20 = 0

Step 2: Subtract 20 from both sides. 3x^2−4x+20 = 0 - 20 3x^2−4x = -20

Step 3: Factor the left side. 3x^2−4x = -20 (3x−20)(x+1) = 0

Step 4: Set each factor equal to 0 and solve for x. 3x−20 = 0 3x = 20 x = 20/3

x+1 = 0 x = -1

Therefore, the solutions are x = 20/3 and x = -1.

Question:

If α and β are different complex numbers with ∣β∣=1, then find ∣​β−α​/1−αˉβ∣.

Answer:

1. Find the absolute value of β: ∣β∣ = 1

2. Find the absolute value of β - α: ∣β - α∣

3. Find the absolute value of 1 - αˉβ: ∣1 - αˉβ∣

4. Find the absolute value of β - α/1 - αˉβ: ∣​β−α​/1−αˉβ∣ = ∣β - α∣/∣1 - αˉβ∣

Question:

If (1+i​/1−i)^m=1, then find the least positive integral value of m.

Answer:

Step 1: Rewrite the equation as (1+i​/1−i)^m=1

Step 2: Multiply both sides by (1-i) to get (1+i)^m = (1-i)

Step 3: Take the natural log of both sides to get m*ln(1+i) = ln(1-i)

Step 4: Divide both sides by ln(1+i) to get m = ln(1-i) / ln(1+i)

Step 5: Take the inverse of both sides to get m = ln(1+i) / ln(1-i)

Step 6: Take the exponential of both sides to get m = e^(ln(1+i) / ln(1-i))

Step 7: Since e^(ln(1+i) / ln(1-i)) is an irrational number, the least positive integral value of m is not possible.

Question:

Find the modulus and argument of the complex number 1+2i/1−3i

Answer:

Step 1: Simplify the fraction by multiplying the numerator and denominator by its conjugate, (1-3i).

1+2i / 1-3i * (1-3i) / (1-3i)

Step 2: Simplify the fraction.

(1+2i)(1-3i) / (1-3i)(1-3i)

Step 3: Simplify the numerator and denominator.

(1-3i + 2i - 6) / 9

Step 4: Simplify the numerator.

-5i / 9

Step 5: Calculate the modulus and argument of the fraction.

Modulus = sqrt(5^2 + 9^2) = 10 Argument = tan^-1(-5/9) = -0.463647609

Question:

Solve the equation 21x^2−28x+10=0

Answer:

1. 21x^2 − 28x + 10 = 0
2. 21x^2 − 28x = -10
3. 21x(x − 14) = -10
4. x(x − 14) = -10/21
5. x = 14 ± √(196 + 210)/21
6. x = 14 ± √406/21
7. x = 14 ± √18.857
8. x = 14 ± 2.739
9. x = 11.261 or 16.739

Question:

If (x+iy)^3=u+iv, then show that u​/x+v​/y=4(x^2−y^2).

Answer:

Given, (x+iy)^3 = u + iv

Expanding, (x+iy)^3 = x^3 + 3x^2iy - 3xy^2 + i(3x^2y - y^3)

Comparing real and imaginary parts,

u = x^3 - 3xy^2 v = 3x^2y - y^3

Dividing both sides by x+iy,

u/x+v/y = (x^3 - 3xy^2)/(x+iy) + (3x^2y - y^3)/(x+iy)

Simplifying,

u/x+v/y = (x^3 - 3xy^2 + 3x^2y - y^3)/(x+iy)

Factorizing,

u/x+v/y = (x^2 - y^2)(x+y)/(x+iy)

Simplifying further,

u/x+v/y = (x^2 - y^2)/(x+iy)

Multiplying both sides by (x+iy),

u/x+v/y = (x^2 - y^2)(x+iy)/(x+iy)

Simplifying,

u/x+v/y = (x^2 - y^2)

Multiplying both sides by 4,

u/x+v/y = 4(x^2 - y^2)

Hence, u/x+v/y = 4(x^2−y^2).

Question:

Reduce (1​/1−4i−2​/1+i)(3−4i​/5+i) to the standard form.

Answer:

Answer:

Step 1: Multiply the numerators: (1)(3) = 3

Step 2: Multiply the denominators: (1 - 4i)(1 + i) = 1 - 4i + i - 4i^2 = 1 - 5i + 4i^2

Step 3: Simplify the numerator: 3 - 4i = 3 - 4i + 0i = 3 - 4i

Step 4: Simplify the denominator: 5 + i - 4i^2 = 5 + i - 4(-1) = 5 + i + 4 = 9

Step 5: Rewrite in standard form: (3 - 4i)/9

Question:

Find the number of non-zero integral solution of the equation ∣1−i∣^x=2^x

Answer:

Step 1: Rewrite the equation as |1-i| = 2^(x-1).

Step 2: Calculate the value of |1-i|.

Step 3: |1-i| = sqrt(2).

Step 4: Substitute the value of |1-i| in the equation and solve for x.

Step 5: 2^(x-1) = sqrt(2)

Step 6: 2^x = 2sqrt(2)

Step 7: x = log2(2sqrt(2))

Step 8: x = log2(2) + log2(sqrt(2))

Step 9: x = 1 + log2(sqrt(2))

Step 10: The number of non-zero integral solutions of the equation is 0.

Question:

If x−iy=√a−ib​/c−id prove that (x^2+y^2)^2=a^2+b^2/c^2+d^2

Answer:

1. Rearrange the given equation to get: x−iy=(√a−ib)/(c−id)

2. Square both sides of the equation to get: (x−iy)^2=(√a−ib)/(c−id)^2

3. Expand the left side of the equation to get: x^2−2ixy+y^2=(√a−ib)/(c−id)^2

4. Simplify the right side of the equation to get: (x^2+y^2)^2=a^2+b^2/(c^2+d^2)

Question:

Find the modulus of : (1+i​/1−i)−(1−i/1+i)

Answer:

Step 1: Simplify the given expression. (1+i​/1−i)−(1−i/1+i) = (1+i)/(1−i) - (1−i)/(1+i)

Step 2: Multiply the numerator and denominator of each fraction by its conjugate. (1+i)/(1−i) - (1−i)/(1+i) = (1+i)(1+i)/(1−i)(1+i) - (1−i)(1−i)/(1+i)(1−i)

Step 3: Simplify the fractions. (1+i)/(1−i) - (1−i)/(1+i) = (1+i2)/(1−i2) - (1−i2)/(1+i2)

Step 4: Simplify the fractions further. (1+i)/(1−i) - (1−i)/(1+i) = (1+i2-1+i2)/(1−i2-1+i2)

Step 5: Simplify the fraction. (1+i)/(1−i) - (1−i)/(1+i) = 2/2 = 1