Modulus of sum is less than or equal to the sum of modulus.

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Given `z_1 = a_1+ib_1` and `z_2 = a_2+ib_2`, what is the modulus of sum `|z_1 + z_2|`?

- `<=|z_1| xx |z_2|`
- `<=|z_1| + |z_2| `
- `<=|z_1| + |z_2| `

The answer is '`<=|z_1| + |z_2|`'.

`|z_1 + z_2| <= |z_1| + |z_2|` is proven with the representation on complex plane. `z_1, z_2, z_1+z_2` form a triangle. Length of the sides of the triangle are `|z_1|, |z_2|, |z_1+z_2|`. Sum of any two sides of the triangle is greater than the third side.

• Modulus of sum is less than or equal to the sum of modulus.

**Modulus of Sum: ** For complex numbers `z_1, z_2 in CC`

`|z_1 + z_2| <= |z_1| + |z_2|`

*Solved Exercise Problem: *

Given `z_1 = a_1+ib_1` and `z_2 = a_2+ib_2` what is the modulus of difference `|z_1 - z_2|`?

- `<=|z_1| - |z_2|`
- `<=|z_1| + |z_2| `
- `<=|z_1| + |z_2| `

The answer is '`<=|z_1| + |z_2|`'. Subtraction is the inverse of addition. So, `|z_1 - z_2|` = `|z_1 + (-z_2)|`. `<= |z_1|+|-z_2|` `<= |z_1| + |z_2|`. Note : `|-z_2| = |z_2|`

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