# Find the absolute value for each complex number: l -4 - 10i l l 3- 6i l

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### 3 Answers

l -4 - 10i l

To find the absolute value for a complex number , we use the following formula:

if z = a+ bi

==> l zl = sqrt(a^2 + b^2)

==> l -4 - 10i l = sqrt(-4^2 + -10^2)

= sqrt(16+100)

= sqrt116)

= 2 sqrt29

**==> l -4 -10i l = 2sqrt29**

l 3- 6i l

==> l 3- 6i l = sqrt(3^2 + -6^2 )

= sqrt(9+ 36)

= sqrt(45)

= 3sqrt5

**==> l 3-6i l = 3sqrt5**

Here we need to find the absolute value for the given complex numbers.

Now for any complex number x + iy, the absolute value is obtained by sqrt (x^2 + y^2)

For -4 - 10i

x= -4 and y = -10, therefore the absolute value is sqrt [(-4) ^2+ (-10) ^2] = sqrt (16 + 100) = sqrt 116 = 10.77 (approximately)

For 3- 6i

x= 3 and y = -6, therefore the absolute value is sqrt [3^2+ (-6) ^2] = sqrt (9 + 36) = sqrt 45 = 6.708 (approximately)

The absolute value of a complex number is the modulus of the complex number.

z = x + i*y

Re(z) = x

Im(z) = y

|z| =sqrt[Re^2(z) + Im^2(z)]

We'll identify x and y for the first number z.

z = -4 - 10i |z| = l -4 - 10i l Re(z) = -4 Im(z) = -10 |z| =sqrt [(-4)^2 + (-10)^2] |z| = sqrt (16 + 100) |z| = sqrt 116**|z| = 2sqrt29**