Question

What is the modulus of |9+40i|?

Answer 1

Answer:  Modulus of |9+40i| =41

Step-by-step explanation:

Since we have given that

$z=\mid9+40i\mid$

Here,

$z=a+bi\\\\where,\\\\a=\text{real number}=9\\\\b=\text{imaginary number}=40$

We need to find the modulus of z:

$\mid z\mid=\sqrt{a^2+b^2}$

$\mid z\mid=\sqrt{9^2+40^2}\\\\\mid z\mid=\sqrt{81+1600}\\\\\mid z\mid=\sqrt{1681}\\\\\mid z\mid=41$

Hence, Modulus of |9+40i| =41

Answer 2

Answer:

41

Step-by-step explanation:

We know that complex numbers are a combination of real and imaginary numbers

Real part is x and imaginary part y is multiplied by i, square root of -1

Modulus of x+iy = $\sqrt{x^2+y^2}$

Here instead of x and y are given 9 and 40

i.e. 9+40i

Hence to find modulus we square the coefficients add them and then find square root

|9+49i| =$\sqrt{9^2+40^2} =\sqrt{1681}$

By long division method we find that

|9+40i| =41