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Zorluk: OrtaComplex Numbers and Operations

For the imaginary unit ii, where i2=−1i^2 = -1, the complex number zz is defined as z=a+bi1−2iz = \frac{a + bi}{1 - 2i}, where aa and bb are real numbers. If z=4+3iz = 4 + 3i, what is the value of a+ba + b?

Cevap: 5

Cevap

The value of a+ba + b is 5.
To find the value of a+ba + b, we start with the equation a+bi1−2i=4+3i\frac{a + bi}{1 - 2i} = 4 + 3i. Multiplying both sides by the denominator gives a+bi=(4+3i)(1−2i)a + bi = (4 + 3i)(1 - 2i). Expanding the right side using the distributive property, we get a+bi=4(1)+4(−2i)+3i(1)+3i(−2i)=4−8i+3i−6i2a + bi = 4(1) + 4(-2i) + 3i(1) + 3i(-2i) = 4 - 8i + 3i - 6i^2. Substituting i2=−1i^2 = -1 simplifies the expression to 4−5i−6(−1)=4−5i+6=10−5i4 - 5i - 6(-1) = 4 - 5i + 6 = 10 - 5i. By comparing the real and imaginary parts of both sides, we find that a=10a = 10 and b=−5b = -5. The sum of these two values is a+b=10+(−5)=5a + b = 10 + (-5) = 5.

Adım Adım Çözüm

1
Isolate the numerator by multiplying both sides by the denominator.
a+bi=(4+3i)(1−2i)a + bi = (4 + 3i)(1 - 2i)
To solve for the variables aa and bb in the numerator, we clear the fraction by multiplying by the denominator.
2
Expand the product of the two complex numbers.
a+bi=4−8i+3i−6i2a + bi = 4 - 8i + 3i - 6i^2
Distribute each term of the first binomial to each term of the second binomial.
3
Simplify the expression using the definition of i2i^2.
a+bi=10−5ia + bi = 10 - 5i
Since i2=−1i^2 = -1, the term −6i2-6i^2 becomes +6+6. Combine the real parts (4+6=104 + 6 = 10) and imaginary parts (−8i+3i=−5i-8i + 3i = -5i).
4
Equate the components and calculate a+ba + b.
a=10a = 10, b=−5b = -5, and a+b=5a + b = 5
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Therefore, a=10a = 10 and b=−5b = -5. Summing these yields 10+(−5)=510 + (-5) = 5.

Anahtar Kavram

Equality and multiplication of complex numbers
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Konular

Complex Numbers and Operations

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