Math / Algebra

QuestionIf $z$ is the solution of the equation: $z-1+i=2 i z$ then $\operatorname{Re}(z)=$ Note: Answer should be in decimal form (do not use fraction)

Studdy Solution

STEP 1

What is this asking? Find the real part of the complex number $z$ that solves the equation $z - 1 + i = 2iz$ . Watch out! Remember that $z$ is a complex number, so we can't just treat it like a regular variable.
We'll need to carefully isolate $z$ and consider its real and imaginary parts separately.

STEP 2

1. Isolate $z$
2. Find the real part of $z$

STEP 3

We start with our given equation: $z - 1 + i = 2iz$ Let's move all the terms with $z$ to one side and the other terms to the other side.
We'll add 1 and subtract $i$ from both sides, and also subtract $2iz$ from both sides: $z - 2iz = 1 - i$

STEP 4

On the left side, we can factor out $z$ : $z(1 - 2i) = 1 - i$

STEP 5

Now, we want to isolate $z$ , so we'll divide both sides by $(1 - 2i)$ : $z = \frac{1 - i}{1 - 2i}$

STEP 6

To simplify this complex fraction, we'll multiply the numerator and denominator by the conjugate of the denominator, which is $(1 + 2i)$ .
This will give us a real number in the denominator: $z = \frac{1 - i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i}$

STEP 7

Expanding the numerator, we get: $(1 - i)(1 + 2i) = 1 + 2i - i - 2i^2 = 1 + i - 2(-1) = 1 + i + 2 = 3 + i$ Expanding the denominator, we get: $(1 - 2i)(1 + 2i) = 1 + 2i - 2i - 4i^2 = 1 - 4(-1) = 1 + 4 = 5$ So, $z = \frac{3 + i}{5}$

STEP 8

We can rewrite this as: $z = \frac{3}{5} + \frac{1}{5}i$

STEP 9

The real part of $z$ is the coefficient of the term without $i$ , which is $\frac{3}{5}$ .
As a decimal, this is 0.6.

STEP 10

The real part of $z$ , or $\operatorname{Re}(z)$ , is 0.6.

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QuestionIf $z$ is the solution of the equation: $z-1+i=2 i z$ then $\operatorname{Re}(z)=$ Note: Answer should be in decimal form (do not use fraction)

Studdy Solution

STEP 1

STEP 2

1. Isolate $z$
2. Find the real part of $z$

STEP 3

We start with our given equation: $z - 1 + i = 2iz$ Let's move all the terms with $z$ to one side and the other terms to the other side.
We'll add 1 and subtract $i$ from both sides, and also subtract $2iz$ from both sides: $z - 2iz = 1 - i$

STEP 4

On the left side, we can factor out $z$ : $z(1 - 2i) = 1 - i$

STEP 5

Now, we want to isolate $z$ , so we'll divide both sides by $(1 - 2i)$ : $z = \frac{1 - i}{1 - 2i}$

STEP 6

To simplify this complex fraction, we'll multiply the numerator and denominator by the conjugate of the denominator, which is $(1 + 2i)$ .
This will give us a real number in the denominator: $z = \frac{1 - i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i}$

STEP 7

Expanding the numerator, we get: $(1 - i)(1 + 2i) = 1 + 2i - i - 2i^2 = 1 + i - 2(-1) = 1 + i + 2 = 3 + i$ Expanding the denominator, we get: $(1 - 2i)(1 + 2i) = 1 + 2i - 2i - 4i^2 = 1 - 4(-1) = 1 + 4 = 5$ So, $z = \frac{3 + i}{5}$

STEP 8

We can rewrite this as: $z = \frac{3}{5} + \frac{1}{5}i$

STEP 9

The real part of $z$ is the coefficient of the term without $i$ , which is $\frac{3}{5}$ .
As a decimal, this is 0.6.

STEP 10

The real part of $z$ , or $\operatorname{Re}(z)$ , is 0.6.

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QuestionIf zzz is the solution of the equation: z−1+i=2izz-1+i=2 i zz−1+i=2iz then Re⁡(z)=\operatorname{Re}(z)=Re(z)= Note: Answer should be in decimal form (do not use fraction)

Studdy Solution

STEP 1

STEP 2

1. Isolate zzz2. Find the real part of zzz

STEP 3

We start with our given equation: z−1+i=2iz z - 1 + i = 2iz z−1+i=2iz Let's **move** all the terms with zzz to one side and the other terms to the other side.We'll add 1 and subtract iii from both sides, and also subtract 2iz2iz2iz from both sides: z−2iz=1−i z - 2iz = 1 - i z−2iz=1−i

STEP 4

On the left side, we can **factor out** zzz: z(1−2i)=1−i z(1 - 2i) = 1 - i z(1−2i)=1−i

STEP 5

Now, we want to **isolate** zzz, so we'll **divide** both sides by (1−2i)(1 - 2i)(1−2i): z=1−i1−2i z = \frac{1 - i}{1 - 2i} z=1−2i1−i​

STEP 6

STEP 7

STEP 8

We can **rewrite** this as: z=35+15i z = \frac{3}{5} + \frac{1}{5}i z=53​+51​i

STEP 9

The real part of zzz is the coefficient of the term without iii, which is 35\frac{3}{5}53​.As a decimal, this is **0.6**.

STEP 10

The real part of zzz, or Re⁡(z)\operatorname{Re}(z)Re(z), is **0.6**.

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QuestionIf $z$ is the solution of the equation: $z-1+i=2 i z$ then $\operatorname{Re}(z)=$ Note: Answer should be in decimal form (do not use fraction)

1. Isolate $z$
2. Find the real part of $z$

We start with our given equation: $z - 1 + i = 2iz$ Let's move all the terms with $z$ to one side and the other terms to the other side.
We'll add 1 and subtract $i$ from both sides, and also subtract $2iz$ from both sides: $z - 2iz = 1 - i$

On the left side, we can factor out $z$ : $z(1 - 2i) = 1 - i$

Now, we want to isolate $z$ , so we'll divide both sides by $(1 - 2i)$ : $z = \frac{1 - i}{1 - 2i}$

We can rewrite this as: $z = \frac{3}{5} + \frac{1}{5}i$

The real part of $z$ is the coefficient of the term without $i$ , which is $\frac{3}{5}$ .
As a decimal, this is 0.6.

The real part of $z$ , or $\operatorname{Re}(z)$ , is 0.6.