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What is the absolute value of z if i(z-1)=-2 ?

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colorcolour | Student, College Freshman | eNoter

Posted December 21, 2010 at 1:54 AM via web

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What is the absolute value of z if i(z-1)=-2 ?

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giorgiana1976 | College Teacher | Valedictorian

Posted December 21, 2010 at 1:57 AM (Answer #1)

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To find the absolute value of the complex number, we'll put it in the rectangular form first.

For this reason, we'll re-write z, isolating z to the left side. For this reason, we'll remove the brackets:

iz - i = - 2

We'll add i both sides:

iz = i - 2

We'll divide by i both sides:

z = (i - 2)/i

Since we have to put z in the rectangular form and since we are not allowed to keep a complex number to the denominator, we'll multiply the ratio by the conjugate of i, that is -i.

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

But i^2 = -1

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

We'll remove the brackets:

z = 2i - i^2

z = 1 + 2i

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

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

|z| = sqrt(1 + 4)

The absolute value of the complex number z is: |z|  = sqrt 5

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 21, 2010 at 1:59 AM (Answer #2)

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First let's determine z from i(z-1)=-2

i(z-1)=-2

=> iz - i = -2

=> iz = i - 2

=> z = i/i - 2/i

=> z = 1 - 2/i

=> z = 1 - 2i / i^2

=> z = 1 + 2i

Now the absolute value is sqrt ( 1^2 + 2^2) = sqrt (1 + 4) = sqrt 5.

The absolute value of z is sqrt 5.

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neela | High School Teacher | Valedictorian

Posted December 21, 2010 at 2:05 AM (Answer #3)

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To find the absolute value of z if i(z-1)=-2

We first solve for z and then determine the absolute value of z.

i(z-1) = -2.

z-1 = -2/i = -2*i/(i*i0 = -2i/(-1) = 2i.

Therefore z -1 = 2i

 z = 2i+1 = 1+2i.

z = 1+2i.

Now absolute value of z= |x+yi| = (x^2+y^2)^(1/2).

|z| = |1+2i| = (1^2+2^2)^(1/2) = 5^(1/2).

So the absolute value of z = 5^(1/2).

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