Graph |z-1+i|<3

$| z - 1 + i | < 3$

Step 1

Simplify $| z - 1 + i |$ .

Use the formula $| a + b i | = \sqrt{a^{2} + b^{2}}$ to find the magnitude.

$\sqrt{{(z - 1)}^{2} + 1^{2}} < 3$

One to any power is one.

$\sqrt{{(z - 1)}^{2} + 1} < 3$

Step 2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{{(z - 1)}^{2} + 1}}^{2} < 3^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{{(z - 1)}^{2} + 1}$ as ${({(z - 1)}^{2} + 1)}^{\frac{1}{2}}$ .

${({({(z - 1)}^{2} + 1)}^{\frac{1}{2}})}^{2} < 3^{2}$

Simplify the left side.

Simplify ${({({(z - 1)}^{2} + 1)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({({(z - 1)}^{2} + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${({(z - 1)}^{2} + 1)}^{\frac{1}{2} \cdot 2} < 3^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${({(z - 1)}^{2} + 1)}^{\frac{1}{2} \cdot 2} < 3^{2}$

Rewrite the expression.

${({(z - 1)}^{2} + 1)}^{1} < 3^{2}$

Simplify.

${(z - 1)}^{2} + 1 < 3^{2}$

Simplify the right side.

Raise $3$ to the power of $2$ .

${(z - 1)}^{2} + 1 < 9$

Step 4

Solve for $z$ .

Subtract $1$ from both sides of the inequality.

${(z - 1)}^{2} < 9 - 1$

Subtract $1$ from $9$ .

${(z - 1)}^{2} < 8$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{{(z - 1)}^{2}} < \sqrt{8}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| z - 1 | < \sqrt{8}$

Simplify the right side.

Simplify $\sqrt{8}$ .

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$| z - 1 | < \sqrt{4 (2)}$

Rewrite $4$ as $2^{2}$ .

$| z - 1 | < \sqrt{2^{2} \cdot 2}$

Pull terms out from under the radical.

$| z - 1 | < | 2 | \sqrt{2}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| z - 1 | < 2 \sqrt{2}$

Write $| z - 1 | < 2 \sqrt{2}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$z - 1 \geq 0$

Add $1$ to both sides of the inequality.

$z \geq 1$

In the piece where $z - 1$ is non-negative, remove the absolute value.

$z - 1 < 2 \sqrt{2}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$z - 1 < 0$

Add $1$ to both sides of the inequality.

$z < 1$

In the piece where $z - 1$ is negative, remove the absolute value and multiply by $- 1$ .

$- (z - 1) < 2 \sqrt{2}$

Write as a piecewise.

${\begin{cases} z - 1 < 2 \sqrt{2} & z \geq 1 \\ - (z - 1) < 2 \sqrt{2} & z < 1 \end{cases}$

Simplify $- (z - 1) < 2 \sqrt{2}$ .

Apply the distributive property.

${\begin{cases} z - 1 < 2 \sqrt{2} & z \geq 1 \\ - z - - 1 < 2 \sqrt{2} & z < 1 \end{cases}$

Multiply $- 1$ by $- 1$ .

${\begin{cases} z - 1 < 2 \sqrt{2} & z \geq 1 \\ - z + 1 < 2 \sqrt{2} & z < 1 \end{cases}$

Solve $z - 1 < 2 \sqrt{2}$ when $z \geq 1$ .

Add $1$ to both sides of the inequality.

$z < 2 \sqrt{2} + 1$

Find the intersection of $z < 2 \sqrt{2} + 1$ and $z \geq 1$ .

$1 \leq z < 2 \sqrt{2} + 1$

Solve $- z + 1 < 2 \sqrt{2}$ when $z < 1$ .

Solve $- z + 1 < 2 \sqrt{2}$ for $z$ .

Subtract $1$ from both sides of the inequality.

$- z < 2 \sqrt{2} - 1$

Divide each term in $- z < 2 \sqrt{2} - 1$ by $- 1$ and simplify.

Divide each term in $- z < 2 \sqrt{2} - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- z}{- 1} > \frac{2 \sqrt{2}}{- 1} + \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{z}{1} > \frac{2 \sqrt{2}}{- 1} + \frac{- 1}{- 1}$

Divide $z$ by $1$ .

$z > \frac{2 \sqrt{2}}{- 1} + \frac{- 1}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{2 \sqrt{2}}{- 1}$ .

$z > - 1 \cdot (2 \sqrt{2}) + \frac{- 1}{- 1}$

Rewrite $- 1 \cdot (2 \sqrt{2})$ as $- (2 \sqrt{2})$ .

$z > - (2 \sqrt{2}) + \frac{- 1}{- 1}$

Multiply $2$ by $- 1$ .

$z > - 2 \sqrt{2} + \frac{- 1}{- 1}$

Divide $- 1$ by $- 1$ .

$z > - 2 \sqrt{2} + 1$

Find the intersection of $z > - 2 \sqrt{2} + 1$ and $z < 1$ .

$- 2 \sqrt{2} + 1 < z < 1$

Find the union of the solutions.

$- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$

Step 5

Use each root to create test intervals.

$z < - 2 \sqrt{2} + 1$

$- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$

$z > 2 \sqrt{2} + 1$

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $z < - 2 \sqrt{2} + 1$ to see if it makes the inequality true.

Choose a value on the interval $z < - 2 \sqrt{2} + 1$ and see if this value makes the original inequality true.

$z = - 4$

Replace $z$ with $- 4$ in the original inequality.

$| (- 4) - 1 + i | < 3$

Test a value on the interval $- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$ to see if it makes the inequality true.

Choose a value on the interval $- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$ and see if this value makes the original inequality true.

$z = 0$

Replace $z$ with $0$ in the original inequality.

$| (0) - 1 + i | < 3$

Test a value on the interval $z > 2 \sqrt{2} + 1$ to see if it makes the inequality true.

Choose a value on the interval $z > 2 \sqrt{2} + 1$ and see if this value makes the original inequality true.

$z = 6$

Replace $z$ with $6$ in the original inequality.

$| (6) - 1 + i | < 3$

Compare the intervals to determine which ones satisfy the original inequality.

$z < - 2 \sqrt{2} + 1$ +|(-4)-1+i|

$- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$ +|(+0)-1+i|

$z > 2 \sqrt{2} + 1$ +|(+6)-1+i|

$z < - 2 \sqrt{2} + 1$ +|(-4)-1+i|

$- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$ +|(+0)-1+i|

$z > 2 \sqrt{2} + 1$ +|(+6)-1+i|

Step 7

The solution consists of all of the true intervals.

$z < - 2 \sqrt{2} + 1$ or $- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$ or $z > 2 \sqrt{2} + 1$

Step 8

\n\n\n

\n\n

\n \n \n \n \n \n \n

\n\n\n\n\n\n

\n\n\t\n \n\n\n\n \n\n\n \n\n\n\n \n\n \n\n\n\n \n\n\n\n \n\n\n\n\n\n\t\n\n\n\n\t\n\n\n\n\n\t\n

\n $\n \n \n \n [\n \begin{array}{c} \n \\ \n & \n \n^{\n x \n \n \n 2 \n \n} \n \n & \n & \n \n \frac{\n}{\n 1 \n} \n \n & \n \\ \n \\ \n & \n \n \sqrt{\n \n π \n \n} \n \n & \n & \n \n \n \n \int \n \n \n \n \n x \n \n d \n \n x \n \n \n \n \n & \n \\ \n \end{array} \n] \n \n \n \n$ \n

\n\n\n\n\n\n\t\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n \n\n \n\t\n \n \n\n\n\t\n\n\n\n\n \n

Step 7

The solution consists of all of the true intervals.

$z < - 2 \sqrt{2} + 1$ or $- 2 \sqrt{2} + 1 < z < 2 \sqrt{2} + 1$ or $z > 2 \sqrt{2} + 1$

Step 8

Graph |z-1+i|<3

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