QuestionFind the modulus of the complex number $z=\frac{2}{1-i}$ .

Studdy Solution

STEP 1

Assumptions1. The complex number is given as $z=\frac{}{1-i}$ . . We need to find the modulus of this complex number.
3. The modulus of a complex number $z=a+bi$ is given by |z|=\sqrt{a^+b^}.

STEP 2

First, we need to express the complex number in the form $a+bi$ .
To do this, we can multiply the numerator and denominator of the fraction by the conjugate of the denominator.
$z=\frac{2}{1-i} \times \frac{1+i}{1+i}$

STEP 3

Now, multiply the numerators and denominators separately.
$z=\frac{2(1+i)}{(1-i)(1+i)}$

STEP 4

implify the denominator using the formula $(a-b)(a+b)=a^2-b^2$ .
$z=\frac{2(1+i)}{1^2-(i)^2}$

STEP 5

implify the denominator further, remembering that $i^2=-1$ .
$z=\frac{2(1+i)}{1-(-1)}$

STEP 6

implify the denominator to get the complex number in the form $a+bi$ .
$z=\frac{2(1+i)}{2}$

STEP 7

implify the complex number.
$z=1+i$

STEP 8

Now that we have the complex number in the form $a+bi$ , we can find the modulus using the formula $|z|=\sqrt{a^2+b^2}$ .
$|z|=\sqrt{(1)^2+(1)^2}$

STEP 9

Calculate the modulus.
$|z|=\sqrt{+}$

STEP 10

Calculate the modulus.
$|z|=\sqrt{2}$ The modulus of $z=\frac{2}{-i}$ is $\sqrt{2}$ .

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QuestionFind the modulus of the complex number $z=\frac{2}{1-i}$ .

Studdy Solution

STEP 1

Assumptions1. The complex number is given as $z=\frac{}{1-i}$ . . We need to find the modulus of this complex number.
3. The modulus of a complex number $z=a+bi$ is given by |z|=\sqrt{a^+b^}.

STEP 2

First, we need to express the complex number in the form $a+bi$ .
To do this, we can multiply the numerator and denominator of the fraction by the conjugate of the denominator.
$z=\frac{2}{1-i} \times \frac{1+i}{1+i}$

STEP 3

Now, multiply the numerators and denominators separately.
$z=\frac{2(1+i)}{(1-i)(1+i)}$

STEP 4

implify the denominator using the formula $(a-b)(a+b)=a^2-b^2$ .
$z=\frac{2(1+i)}{1^2-(i)^2}$

STEP 5

implify the denominator further, remembering that $i^2=-1$ .
$z=\frac{2(1+i)}{1-(-1)}$

STEP 6

implify the denominator to get the complex number in the form $a+bi$ .
$z=\frac{2(1+i)}{2}$

STEP 7

implify the complex number.
$z=1+i$

STEP 8

Now that we have the complex number in the form $a+bi$ , we can find the modulus using the formula $|z|=\sqrt{a^2+b^2}$ .
$|z|=\sqrt{(1)^2+(1)^2}$

STEP 9

Calculate the modulus.
$|z|=\sqrt{+}$

STEP 10

Calculate the modulus.
$|z|=\sqrt{2}$ The modulus of $z=\frac{2}{-i}$ is $\sqrt{2}$ .

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QuestionFind the modulus of the complex number z=21−iz=\frac{2}{1-i}z=1−i2​.

Studdy Solution

STEP 1

Assumptions1. The complex number is given as z=1−iz=\frac{}{1-i}z=1−i​. . We need to find the modulus of this complex number.3. The modulus of a complex number z=a+biz=a+biz=a+bi is given by |z|=\sqrt{a^+b^}.

STEP 2

First, we need to express the complex number in the form a+bia+bia+bi.To do this, we can multiply the numerator and denominator of the fraction by the conjugate of the denominator.z=21−i×1+i1+iz=\frac{2}{1-i} \times \frac{1+i}{1+i}z=1−i2​×1+i1+i​

STEP 3

Now, multiply the numerators and denominators separately.z=2(1+i)(1−i)(1+i)z=\frac{2(1+i)}{(1-i)(1+i)}z=(1−i)(1+i)2(1+i)​

STEP 4

implify the denominator using the formula (a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2(a−b)(a+b)=a2−b2.z=2(1+i)12−(i)2z=\frac{2(1+i)}{1^2-(i)^2}z=12−(i)22(1+i)​

STEP 5

implify the denominator further, remembering that i2=−1i^2=-1i2=−1.z=2(1+i)1−(−1)z=\frac{2(1+i)}{1-(-1)}z=1−(−1)2(1+i)​

STEP 6

implify the denominator to get the complex number in the form a+bia+bia+bi.z=2(1+i)2z=\frac{2(1+i)}{2}z=22(1+i)​

STEP 7

implify the complex number.z=1+iz=1+iz=1+i

STEP 8

Now that we have the complex number in the form a+bia+bia+bi, we can find the modulus using the formula ∣z∣=a2+b2|z|=\sqrt{a^2+b^2}∣z∣=a2+b2​.∣z∣=(1)2+(1)2|z|=\sqrt{(1)^2+(1)^2}∣z∣=(1)2+(1)2​

STEP 9

Calculate the modulus.∣z∣=+|z|=\sqrt{+}∣z∣=+​

STEP 10

Calculate the modulus.∣z∣=2|z|=\sqrt{2}∣z∣=2​The modulus of z=2−iz=\frac{2}{-i}z=−i2​ is 2\sqrt{2}2​.

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QuestionFind the modulus of the complex number $z=\frac{2}{1-i}$ .

Assumptions1. The complex number is given as $z=\frac{}{1-i}$ . . We need to find the modulus of this complex number.
3. The modulus of a complex number $z=a+bi$ is given by |z|=\sqrt{a^+b^}.

First, we need to express the complex number in the form $a+bi$ .
To do this, we can multiply the numerator and denominator of the fraction by the conjugate of the denominator.
$z=\frac{2}{1-i} \times \frac{1+i}{1+i}$

Now, multiply the numerators and denominators separately.
$z=\frac{2(1+i)}{(1-i)(1+i)}$

implify the denominator using the formula $(a-b)(a+b)=a^2-b^2$ .
$z=\frac{2(1+i)}{1^2-(i)^2}$

implify the denominator further, remembering that $i^2=-1$ .
$z=\frac{2(1+i)}{1-(-1)}$

implify the denominator to get the complex number in the form $a+bi$ .
$z=\frac{2(1+i)}{2}$

implify the complex number.
$z=1+i$

Now that we have the complex number in the form $a+bi$ , we can find the modulus using the formula $|z|=\sqrt{a^2+b^2}$ .
$|z|=\sqrt{(1)^2+(1)^2}$

Calculate the modulus.
$|z|=\sqrt{+}$

Calculate the modulus.
$|z|=\sqrt{2}$ The modulus of $z=\frac{2}{-i}$ is $\sqrt{2}$ .