Numerical value of complex number

[tomwilliam]

Homework Statement

I need to understand why
$$\left|4+i \right|=4.123$$

and why this is shown by:

$$\sqrt{4^{2}+1^{2}}=4.123$$

Homework Equations

$$i^{2}=-1$$

The Attempt at a Solution

If I find the square root of this expression squared, then I come up with
$$\sqrt{16-1+8i}$$ which is wrong...

 

[Mentallic]

If you look at an argand diagram, the modulus or |z| of a complex number is the distance from the origin to the complex number. So it's quite simply pythagoras' theorem applied to the problem.

 

[tiny-tim]

hi tomwilliam!

the modulus squared, |a + bi|² is the square of a real number …

it is not the square of any complex number …

|a + bi|² = (a + bi)(a - bi)

 

[tomwilliam]

Thanks. So if I think of the numbers (4, 0) and (0, i) as position vectors on the Argand diagram, where the x-axis is the real number scale and the y-axis is the imaginary number scale, and accept that the modulus is equal to the magnitude of the resultant vector, this means I can ignore the sign of the imaginary part?
Tiny-tim...does the formula you state hold for the modulus squared of any complex number? I would have imagined it was (a+bi)(a+bi)...

 

[tiny-tim]

hi tomwilliam!

… this means I can ignore the sign of the imaginary part?

yes … i² and (-i)² are the same

Tiny-tim...does the formula you state hold for the modulus squared of any complex number? I would have imagined it was (a+bi)(a+bi)...

erm …

my formula is right, your formula is wrong ​

 

[tomwilliam]

Ok, thanks for your help.

 

[Mentallic]

Squaring is not the same as finding the modulus of a complex number.

For any complex number z=a+ib where a and b are real numbers, the modulus is $$|z|=\sqrt{a^2+b^2}$$ which is obviously always positive since a real number squared is positive, thus a²>0, b²>0 then a²+b²>0, and taking the square root is positive. And this makes sense because we can only have a positive length of a vector.
So yes you can ignore the sign on an imaginary number. $$|a+ib|=|-a+ib|=|a-ib|=|-a-ib|$$.

$$z^2=a^2+b^2+2abi$$ which itself is also a complex number, unless a or b are 0.

 

[HallsofIvy]

The modulus of a+ bi is $\sqrt{a^2+ b^2}= \sqrt{(a+ bi)(a- bi)}$.

More generally, the modulus of the complex number, z, is NOT $\sqrt{z^2}$ because $z^2$ is not, in general, a positive real number. It is, rather, $\sqrt{z\overline{z}}$.