Question

find the modules of the complex number (4+5i)(5-4i)​

Answer 1

Given complex number,

$z=(4+5i)(5-4i)$

${\implies z = 4(5-4i) + 5i(5 - 4i)}$

${\implies z = 20-16i + 25i - 20 {i}^{2}}$

${\implies z = 20 + 9i - 20( - 1)}$

${\implies z = 20 + 9i + 20}$

$\boxed{\implies z = 40 + 9i }$

Now, modulus of a complex number is given by,

$|z| = \sqrt{ {x}^{2} + {y}^{2} }$

Here,

• x = real part of z = 40
• y = imaginary part of z = 9

By substituting the values, we get:

$\implies |z| = \sqrt{ {(40)}^{2} + {(9)}^{2} }$

$\implies |z| = \sqrt{ 1600 + 81 }$

$\implies |z| = \sqrt{ 1681 }$

$\implies |z| = 41$

Therefore, the required modulus of the given complex number is 41.

Additional Information:-

• i = √-1
• i² = -1
• i³ = -i
• i⁴ = 1

Whenever, we are given iota raise to the power multiple of 4, the answer will be 1 every time.

Let's say we are asked to find the value of i¹⁰⁰⁰, now since 1000 is divisible by 4, value of i¹⁰⁰⁰ = 1.