$1-\mathrm{i}$

Project ID: 
3000000018
Question: 

Eva had to find the trigonometric form of the complex number $1-\mathrm{i}$.

She proceeded as follows:

(1) First, she calculated the modulus of the given complex number: $$|1-\mathrm{i}|=\sqrt2$$

(2) She remembered from school that a complex number in trigonometric form has its modulus factored out from the parentheses. So she took the given number and factored out $\sqrt2$ from it: $$1-\mathrm{i}=\sqrt2\left(\frac{1}{\sqrt2}-\frac{\mathrm{i}}{\sqrt2}\right)$$

(3) Then she rationalized the denominators and obtained the complex number in the form: $$\sqrt2\left(\frac{\sqrt2}{2}-\mathrm{i}\frac{\sqrt2}{2}\right)$$

(4) Finally, she wrote the expression in parentheses using trigonometric functions and declared that this was the trigonometric form of the given complex number: $$\sqrt2\left(\cos\frac{\pi}{4}-\mathrm{i}\sin\frac{\pi}{4}\right)$$

In which step of her solution did Eva make a mistake?

Answer 1: 

In step (1). The correct calculation of the absolute value should be: $$|1-\mathrm{i}|=\sqrt{1^2+(-\mathrm{i})^2}=0$$

Answer 2: 

In step (2). Correctly, it should be:
$$1-\mathrm{i}=\sqrt2\left(\frac{1}{\sqrt2}-\mathrm{i}\right)$$

Answer 3: 

In step (3). Correctly, it should be: $$\sqrt2(\sqrt2-\mathrm{i}\sqrt2)$$

Answer 4: 

In step (4). The given result is not the trigonometric form of the given complex number.

Fixed Answer: 
All Fixed
Correct Answer: 
Answer 4
Hint: 

The trigonometric form must contain the sum, not the difference, i.e. $$z = r \left(\cos\varphi +\sin\varphi \right)$$

Therefore, it is necessary to find an argument $\varphi$ for which the following holds: $$\cos\varphi=\frac{\sqrt2}{2}\ \wedge\ \sin\varphi=-\frac{\sqrt2}{2}$$

The solution is therefore: $$\varphi=\frac{7\pi}{4}+2k\pi;\ k\in\mathbb{Z}$$

The trigonometric form of the given complex number is, for example: $$1-\mathrm{i}=\sqrt2\left(\cos\frac{7\pi}{4}+\mathrm{i}\sin\frac{7\pi}{4}\right)$$