How do you write 5-i in trigonometric form

Thank you! If you could teach me, that would be greatly appreciated!

Let's say p is the magnitude (or length or modulus) of the number. p = sqrt(5^2 + (-i)^2) = sqrt(25 + 1) = sqrt(26).

Now we need to find the angle of the number with the positive real axis. 5 tells us that part of the triangle is 5 units right, and -i tells us that part of the triangle is 1 unit down. Now, we know that tan(x) = opposite/adjacent. The -1 is opposite and the 5 is adjacent. Hence, tan(x) = -1/5. Therefore, x = arctan(-1/5). Since arctan(-1/5) is in the 2nd quadrant instead of the 4th quadrant, we need to add pi.

So, x = arctan(-1/5) + pi.

No, complex numbers are written in the format p(cos(x) + i sin(x))

So we get 5 - i = sqrt(26) [cos(arctan(-1/5) + pi) + i sin(arctan(-1/5) + pi)]