Examkin

Complex Numbers and Operations

25 soru

Tüm türler25Çoktan seçmeli14Sayısal11
Soru 21Soru →

Let the complex number zz be defined as z=(4−3i)(1+2i)+5i14z = (4 - 3i)(1 + 2i) + 5i^{14}, where i=−1i = \sqrt{-1}. What is the real part of zz?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The real part of the complex number zz is 55.
First, expand the product (4−3i)(1+2i)(4 - 3i)(1 + 2i) to get 4+8i−3i−6i24 + 8i - 3i - 6i^2. Replacing i2i^2 with −1-1 gives 10+5i10 + 5i. Next, simplify 5i145i^{14}. Since i14=(i4)3⋅i2=13⋅(−1)=−1i^{14} = (i^4)^3 \cdot i^2 = 1^3 \cdot (-1) = -1, the term becomes −5-5. Adding the components together gives z=(10+5i)−5=5+5iz = (10 + 5i) - 5 = 5 + 5i. The real part of this complex number is 55.

Adım Adım Çözüm

1
Expand the product of the complex binomials (4−3i)(1+2i)(4 - 3i)(1 + 2i)
10 + 5i
Applying the distributive property gives 4+8i−3i−6i24 + 8i - 3i - 6i^2. Substituting i2=−1i^2 = -1 simplifies the expression to 4+5i+6=10+5i4 + 5i + 6 = 10 + 5i.
2
Simplify the power of the imaginary unit in 5i145i^{14}
-5
Since the powers of ii cycle every 4 terms, i14=i12⋅i2=1⋅(−1)=−1i^{14} = i^{12} \cdot i^2 = 1 \cdot (-1) = -1. Therefore, 5i14=5(−1)=−55i^{14} = 5(-1) = -5.
3
Add the simplified terms together to find zz
5 + 5i
Adding the real and imaginary parts of the terms yields z=(10+5i)+(−5)=5+5iz = (10 + 5i) + (-5) = 5 + 5i.
4
Determine the real part of zz
5
A complex number is written in the form a+bia + bi, where aa represents the real part. For 5+5i5 + 5i, the real part is 55.

Anahtar Kavram

Complex multiplication and simplification of powers of the imaginary unit
Soru 22Soru →

For the imaginary unit ii, where i2=−1i^2 = -1, and any integer nn, what is the value of the expression i4n+3(1+i)8(1−i)6\frac{i^{4n+3} (1 + i)^8}{(1 - i)^6}?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

−2-2
To find the value of the expression, we simplify each part. First, i4n+3i^{4n+3} simplifies to −i-i because i4n=1i^{4n} = 1 and i3=−ii^3 = -i. Second, (1+i)8(1+i)^8 is simplified by first squaring the base to get (1+i)2=2i(1+i)^2 = 2i, and then raising the result to the fourth power: (2i)4=16(2i)^4 = 16. Third, (1−i)6(1-i)^6 is simplified by first squaring the base to get (1−i)2=−2i(1-i)^2 = -2i, and then cubing the result: (−2i)3=−8i3=8i(-2i)^3 = -8i^3 = 8i. Substituting these back into the expression yields −i⋅168i=−2\frac{-i \cdot 16}{8i} = -2. Thus, the expression simplifies to −2-2.

Adım Adım Çözüm

1
Simplify the term i4n+3i^{4n+3} using the properties of powers of ii.
i4n+3=−ii^{4n+3} = -i
Since i4=1i^4 = 1, we can rewrite i4n+3i^{4n+3} as (i4)n⋅i3=1n⋅(−i)=−i(i^4)^n \cdot i^3 = 1^n \cdot (-i) = -i.
2
Simplify the numerator term (1+i)8(1+i)^8.
(1+i)8=16(1+i)^8 = 16
We can rewrite (1+i)8(1+i)^8 as ((1+i)2)4((1+i)^2)^4. Since (1+i)2=1+2i+i2=2i(1+i)^2 = 1 + 2i + i^2 = 2i, we have (2i)4=24⋅i4=16⋅1=16(2i)^4 = 2^4 \cdot i^4 = 16 \cdot 1 = 16.
3
Simplify the denominator term (1−i)6(1-i)^6.
(1−i)6=8i(1-i)^6 = 8i
We can rewrite (1−i)6(1-i)^6 as ((1−i)2)3((1-i)^2)^3. Since (1−i)2=1−2i+i2=−2i(1-i)^2 = 1 - 2i + i^2 = -2i, we have (−2i)3=(−2)3⋅i3=−8⋅(−i)=8i(-2i)^3 = (-2)^3 \cdot i^3 = -8 \cdot (-i) = 8i.
4
Substitute the simplified components back into the original expression and divide.
−i⋅168i=−2\frac{-i \cdot 16}{8i} = -2
Substituting the terms gives −i⋅168i\frac{-i \cdot 16}{8i}. The common factor of ii in the numerator and denominator cancels out, and dividing −16-16 by 88 yields −2-2.

Anahtar Kavram

Simplifying complex expressions involving powers of the imaginary unit and powers of complex binomials.
Tahmini Süre:1m 30s
Soru 23Soru →

For the imaginary unit ii, where i2=−1i^2 = -1, the complex number zz is defined as z=(1+2i)32−iz = \frac{(1 + 2i)^3}{2 - i}. What is the absolute value of zz?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5
The complex number zz simplifies to −4−3i-4 - 3i. The absolute value of a complex number a+bia + bi is defined as a2+b2\sqrt{a^2 + b^2}, which gives (−4)2+(−3)2=16+9=5\sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5. Alternatively, using properties of the modulus, ∣z∣=∣1+2i∣3∣2−i∣=(12+22)322+(−1)2=(5)35=5|z| = \frac{|1 + 2i|^3}{|2 - i|} = \frac{(\sqrt{1^2 + 2^2})^3}{\sqrt{2^2 + (-1)^2}} = \frac{(\sqrt{5})^3}{\sqrt{5}} = 5.

Adım Adım Çözüm

1
Expand the numerator (1+2i)3(1 + 2i)^3
-11 - 2i
To begin simplifying the complex fraction by expanding the cubed binomial in the numerator.
2
Divide the result by the denominator 2−i2 - i by multiplying both numerator and denominator by the complex conjugate 2+i2 + i
-4 - 3i
To express the quotient in standard complex form a+bia + bi.
3
Find the absolute value of the simplified complex number z=−4−3iz = -4 - 3i
5
To calculate the final magnitude of the complex number using ∣a+bi∣=a2+b2|a + bi| = \sqrt{a^2 + b^2}.

Anahtar Kavram

Complex number operations, complex division, and absolute value (modulus) of complex numbers
Soru 24Soru →

If ii represents the imaginary unit, and the complex number zz satisfies the equation z(2+i)−5i97=253−4iz(2 + i) - 5i^{97} = \frac{25}{3 - 4i}, what is the value of z2z^2?

Cevabı ve açıklamayı göster

Cevap: 18i

Cevap

18i
Simplifying the original equation yields the complex number z=3+3iz = 3 + 3i. Squaring this number gives (3+3i)2=9+18i+9i2=9+18i−9=18i(3 + 3i)^2 = 9 + 18i + 9i^2 = 9 + 18i - 9 = 18i.

Adım Adım Çözüm

1
Simplify the fraction on the right side of the equation by multiplying the numerator and denominator by the complex conjugate of the denominator, 3+4i3 + 4i.
The fraction simplifies to 3+4i3 + 4i.
Multiplying by the conjugate rationalizes the denominator: 253−4i=25(3+4i)(3−4i)(3+4i)=25(3+4i)9+16=3+4i\frac{25}{3 - 4i} = \frac{25(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{25(3 + 4i)}{9 + 16} = 3 + 4i.
2
Simplify the power of the imaginary unit, i97i^{97}, by dividing the exponent by 4 to find the remainder.
5i97=5i5i^{97} = 5i
Since 97=4×24+197 = 4 \times 24 + 1, the expression simplifies as i97=(i4)24⋅i=124⋅i=ii^{97} = (i^4)^{24} \cdot i = 1^{24} \cdot i = i.
3
Substitute the simplified expressions back into the original equation and isolate the term containing zz.
z(2+i)=3+9iz(2 + i) = 3 + 9i
Substituting gives z(2+i)−5i=3+4iz(2 + i) - 5i = 3 + 4i. Adding 5i5i to both sides yields z(2+i)=3+9iz(2 + i) = 3 + 9i.
4
Solve for zz by dividing both sides by 2+i2 + i, then simplify by multiplying by the conjugate of the denominator, 2−i2 - i.
z=3+3iz = 3 + 3i
Performing the division: z=3+9i2+i=(3+9i)(2−i)(2+i)(2−i)=6−3i+18i−9i24−i2=15+15i5=3+3iz = \frac{3 + 9i}{2 + i} = \frac{(3 + 9i)(2 - i)}{(2 + i)(2 - i)} = \frac{6 - 3i + 18i - 9i^2}{4 - i^2} = \frac{15 + 15i}{5} = 3 + 3i.
5
Calculate the value of z2z^2 by squaring the complex number 3+3i3 + 3i.
z2=18iz^2 = 18i
Squaring the binomial gives (3+3i)2=9+18i+9i2=9+18i−9=18i(3 + 3i)^2 = 9 + 18i + 9i^2 = 9 + 18i - 9 = 18i.

Anahtar Kavram

Solving equations containing complex numbers by performing operations such as multiplication, division using complex conjugates, and simplifying powers of ii.
Tahmini Süre:2m 30s
Soru 25Soru →

For the imaginary unit ii, where i2=−1i^2 = -1, which of the following complex numbers is equal to 8+i3+2i\frac{8 + i}{3 + 2i}?

Cevabı ve açıklamayı göster

Cevap: 2 - i

Cevap

The complex number 2−i2 - i
To divide two complex numbers, we multiply both the numerator and denominator by the complex conjugate of the denominator, which is 3−2i3 - 2i. Expanding the numerator gives (8+i)(3−2i)=24−16i+3i−2i2=26−13i(8 + i)(3 - 2i) = 24 - 16i + 3i - 2i^2 = 26 - 13i because i2=−1i^2 = -1. Expanding the denominator yields (3+2i)(3−2i)=9−4i2=9+4=13(3 + 2i)(3 - 2i) = 9 - 4i^2 = 9 + 4 = 13. Dividing the terms of the numerator by the denominator gives 26−13i13=2−i\frac{26 - 13i}{13} = 2 - i.

Adım Adım Çözüm

1
Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator.
8+i3+2i⋅3−2i3−2i=(8+i)(3−2i)(3+2i)(3−2i)\frac{8 + i}{3 + 2i} \cdot \frac{3 - 2i}{3 - 2i} = \frac{(8 + i)(3 - 2i)}{(3 + 2i)(3 - 2i)}
Multiplying by the conjugate rationalizes the denominator, converting it into a real number.
2
Expand the numerator and the denominator using binomial multiplication.
Numerator: (8+i)(3−2i)=24−16i+3i−2i2(8 + i)(3 - 2i) = 24 - 16i + 3i - 2i^2
Denominator: (3+2i)(3−2i)=9−6i+6i−4i2=9−4i2(3 + 2i)(3 - 2i) = 9 - 6i + 6i - 4i^2 = 9 - 4i^2
Distribute each term in the first binomial to each term in the second binomial.
3
Substitute i2=−1i^2 = -1 and simplify both expressions.
Numerator: 24−13i−2(−1)=24−13i+2=26−13i24 - 13i - 2(-1) = 24 - 13i + 2 = 26 - 13i
Denominator: 9−4(−1)=9+4=139 - 4(-1) = 9 + 4 = 13
The definition of the imaginary unit is i2=−1i^2 = -1.
4
Divide each term of the simplified numerator by the simplified denominator.
26−13i13=2613−13i13=2−i\frac{26 - 13i}{13} = \frac{26}{13} - \frac{13i}{13} = 2 - i
Separate the real and imaginary parts to write the complex number in standard form a+bia + bi.

Anahtar Kavram

Division of complex numbers using the complex conjugate of the denominator.

Alternatif Yöntem

Instead of dividing directly, let the result be x+yix + yi. Then (x+yi)(3+2i)=8+i(x + yi)(3 + 2i) = 8 + i. Expanding this gives (3x−2y)+(2x+3y)i=8+i(3x - 2y) + (2x + 3y)i = 8 + i. Equating the real and imaginary parts gives the system of equations 3x−2y=83x - 2y = 8 and 2x+3y=12x + 3y = 1. Solving this system yields x=2x = 2 and y=−1y = -1, which corresponds to the complex number 2−i2 - i.
Tahmini Süre:1m 30s
ÖncekiSayfa 2 / 2
Complex Numbers and Operations Alıştırma Soruları — ACT — Sayfa 2 | Examkin