Question

Find the modulus of (1-i/3+i)+(4i/5)

Answer 1

Answer:

your answer is √5/2

Find the modulus of (1-i/3+i)+(4i/5)

Answer 2

Final Answer:

The modulus of the expression $\frac{1-i}{3+i}+\frac{4i}{5}$ is ${\frac{1}{\sqrt5}}$.

Given:

The expression $\frac{1-i}{3+i}+\frac{4i}{5}$ is provided.

To Find:

The modulus of the expression $\frac{1-i}{3+i}+\frac{4i}{5}$ is to be calculated.

Explanation:

The concepts that are important to lead to the solution to the present problem are as follows

• The complex number has two parts, one is real and the other is imaginary.
• Every complex number has its conjugate with the sign of the imaginary part changed from plus sign to minus sign and vice versa.

• The product of the complex number and its conjugate gives a real number.

• The modulus of the complex number $z=x+iy$ is $|z|=\sqrt{x^2+y^2}$

Step 1 of 3

Simplify the complex number given in the problem in the following way.

$\frac{1-i}{3+i}+\frac{4i}{5}\\=\frac{(1-i)(3-i)}{(3+i)(3-i)}+\frac{4i}{5}\\=\frac{1\times (3-i)-i\times (3-i)}{3^2-i^2}+\frac{4i}{5}\\\\=\frac{3-i-3i+i^2}{3^2-i^2}+\frac{4i}{5}$

Step 2 of 3

Using the information $i^2=-1$ in the given problem, simplify the above expression further in the following way.

$\frac{3-i-3i+i^2}{3^2-i^2}+\frac{4i}{5}\\\\=\frac{3-i-3i-1}{9+1}+\frac{4i}{5}\\\\=\frac{2-4i}{10}+\frac{4i}{5}\\\\=\frac{2-4i+2\times 4i}{10}\\\\=\frac{2-4i+8i}{10}\\\\=\frac{2+4i}{10}\\\\=\frac{1+2i}{5}$

Step 3 of 3

In continuation with the above calculations, the  modulus of the expression $\frac{1-i}{3+i}+\frac{4i}{5}$ is

$|\frac{1-i}{3+i}+\frac{4i}{5}|\\=|\frac{1+2i}{5}|\\=|\frac{1}{5}+\frac{2i}{5}|\\=\sqrt{(\frac{1}{5})^2+(\frac{2}{5})^2}\\=\sqrt{(\frac{1}{25})+(\frac{4}{25})}\\=\sqrt{(\frac{1 +4}{25})}\\=\sqrt{(\frac{5}{25})}\\={\frac{1}{\sqrt5}}$

Therefore, the required correct answer is the modulus ${\frac{1}{\sqrt5}}$.

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