Complex Numbers and Quadratic Equations Miscellaneous Exercise
Question:
Find the real numbers x and y , if (x−iy)(3+5i) is the conjugate of −6−24i.
Answer:

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.

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
 Now, we can set up two equations with two unknowns, x and y, to solve for these values:
3x−5y = −6 5x+3y = 24
 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
 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:

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

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

Simplify: x = (10 ± 2√3)/54

Solve: x = (10 + 2√3)/54 or x = (10  2√3)/54
Question:
For any two complex numbers z1 and z2, prove that Re(z1z2)=Rez1Rez2−Imz1Imz2
Answer:
Given: z1 = a + ib, z2 = c + id

Re(z1z2) = Re(ac + ibc + iad  bd)

Re(z1z2) = ac  bd

Re(z1z2) = Re(z1)Re(z2)  Im(z1)Im(z2)

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

Re(z1z2) = ac + bcd + iad  bd

Re(z1z2) = ac  bd

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)/((2i)(2+i)) = (1+7i)/(4+2i2i1) = (1+7i)/(3+i) = (1/3+7i/3)/(1+i) = (1/3+7i/3)(1i)/(1+i)(1i) = (1/3+7i/3)(1i)/(1+i^2) = (1/3+7i/3)(1i)/2 = (1/6+7i/6)/(1+i)
Polar form: (1/6+7i/6)∠tan^1(1/1)
(ii) 1+3i/1−2i = (1+3i)/(12i) = (1+3i)(1+2i)/(12i)(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(z1z2/z1ˉ) (ii) Im(1/z1z1ˉ)
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)(acegacehib+adbceiadbdfi^2+bcegfibcehi^2+bdfegibdfhi^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)(acegacehib+adbceiadbdfi^2+bcegfibcehi^2+bdfegibdfhi^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:

Find the absolute value of β: ∣β∣ = 1

Find the absolute value of β  α: ∣β  α∣

Find the absolute value of 1  αˉβ: ∣1  αˉβ∣

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 (1i) to get (1+i)^m = (1i)
Step 3: Take the natural log of both sides to get m*ln(1+i) = ln(1i)
Step 4: Divide both sides by ln(1+i) to get m = ln(1i) / ln(1+i)
Step 5: Take the inverse of both sides to get m = ln(1+i) / ln(1i)
Step 6: Take the exponential of both sides to get m = e^(ln(1+i) / ln(1i))
Step 7: Since e^(ln(1+i) / ln(1i)) 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, (13i).
1+2i / 13i * (13i) / (13i)
Step 2: Simplify the fraction.
(1+2i)(13i) / (13i)(13i)
Step 3: Simplify the numerator and denominator.
(13i + 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:
 21x^2 − 28x + 10 = 0
 21x^2 − 28x = 10
 21x(x − 14) = 10
 x(x − 14) = 10/21
 x = 14 ± √(196 + 210)/21
 x = 14 ± √406/21
 x = 14 ± √18.857
 x = 14 ± 2.739
 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 nonzero integral solution of the equation ∣1−i∣^x=2^x
Answer:
Step 1: Rewrite the equation as 1i = 2^(x1).
Step 2: Calculate the value of 1i.
Step 3: 1i = sqrt(2).
Step 4: Substitute the value of 1i in the equation and solve for x.
Step 5: 2^(x1) = 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 nonzero 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:

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

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

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

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+i21+i2)/(1−i21+i2)
Step 5: Simplify the fraction. (1+i)/(1−i)  (1−i)/(1+i) = 2/2 = 1