# What is |z| if i - iz = 2 ?

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### 3 Answers

i - iz = 2

First we need to rewrite the function as a function oin the following format:

z = a + bi

Let us isolate z on the left side:

We need to subtract i from both sides:

==> -zi = 2 - i

Now we will divide by -i :

==> z - (2-i)/ -i

Now we will mulitply and divide by i :

==> z = (2-i)*i / -i*i

= (2i - i^2)/ -i^2

But we know that:

i = sqrt-1 ==> i^2 = -1 ==> -i^2 = 1

==? z = (2i + 1)/ 1

==? z = 2i + 1

**==> z = 1 + 2i**

**l zl = sqrt( 1^2 + 2^2)**

** = sqrt(5) **

**==> lzl = sqrt5**

Given i-iz = 2.

To find the value of |z|.

We know that |x+yi| = sqrt(x^2+y^2).

Since i-iz = 2,

-iz = 2-i.

z = 2-i/-i

z = -(2-i)i/i*i , we multiplied both numerator and denomiator by i.

z = (-2i+i^2)/i^2.

z = (-2i+1)/(-1), as i^2 = 1.

z = (-1 +2i).

Therefore |z| = sqrt{(-1)^2+(2)^2} = sqrt(5).

Therefore |z| = sqrt5.

We'll re-write z, isolating z to the left side. For this reason, we'll subtract i both sides:

i - iz = 2

-iz = 2 - i

We'll divide by -i:

z = (2-i)/-i

Since we have to put z in the rectangular form:

z = x + i*y, we'll multiply the ratio by the conjugate of -i, that is i.

z = i*(2-i)/-i^2

But i^2 = -1

z = i*(2-i)/-(-1)

We'll remove the brackets:

z = 2i - i^2

z = 1 + 2i

gThe modulus of z: |z| = sqrt (x^2 + y^2)

We'll identify x = 1 and y = 2.

|z| = sqrt(1 + 4)

**|z| =sqrt 5**