How to write a complex number in the form $re^{i \phi}$ when $r \ge 0$

Question

$$\frac{1+i}{1-i}$$and$$(1+i)(1-i).$$

The answers are $e^{\frac{i\pi}{2}}$ and $2e^{i\phi}$.

Answer 1

welcome to MSE! Please, next time try to explain better what is the problem and what you tried to do to solve it, possibly using MathJax.

For the first fraction, first it is better to reduce it: $$\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1+i)(1-i)}=\frac{2i}2=i.$$ Then, it is simple to see that the number $r=i$ has radius $1$ and argument $\theta=\pi/2,$ cause it can be written as $0+1\cdot i,$ so $r=\sqrt{0^2+1^2}=1$ and $\theta=cotan^{-1}(0/1)=\pi/2.$

For the product, as just said $(i+1)(1-i)=2$ and in this case the radius is equal to $2$ and the argument is $0.$

Answer 2

Hint:

$$\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{2i}{2}=i$$

If you plot this point in the complex plane, can you find $\phi$ and $r$?

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For instance, if you were to plot $z=\frac{1+i}{2}$ you would get something like:

Answer 3

$(1+i)(1-i)=1-i^2=1+1=2$

So if you are going to write this as $2e^{i \phi}$ then $e^{i \phi}=1$. So $\phi=2n\pi$ where $n\in \mathbb{Z}$