Express $$\sqrt {\dfrac {1+i}{1-i}}$$ in the form of $a+ib$.
My Attempt: $$\sqrt {\dfrac {1+i}{1-i}} =\sqrt {\dfrac {1+i}{1-i} \times \dfrac {1+i}{1+i}}=\sqrt {i}.$$
$\begingroup$
$\endgroup$
1
-
$\begingroup$ @quanticbolt, expand $(1+i)^2$ and simplify.. $\endgroup$– pi-πCommented Nov 16, 2017 at 15:48
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
Note that $$\frac {1+i}{1-i}=\frac {(1+i)(1+i)}{(1-i)(1+i)}=\frac {(1+i)^2}{1-i+i-i^2}=\frac {(1+i)^2}{2}\quad\left(=\frac {1+2i+i^2}{2}=i\right).$$ (the further simplification to $i$ does not help you here). Hence the TWO square roots are $$\frac{1+i}{\sqrt{2}}\quad \text{and}\quad -\frac{1+i}{\sqrt{2}}.$$
answered Nov 16, 2017 at 15:51
-
1$\begingroup$ To be fair, those are the two square roots of $i.$ The main error in the OP's attempt is that $\sqrt{i}$ is not in the form $a+ib.$ $\endgroup$– David KCommented Nov 16, 2017 at 15:53
-
$\begingroup$
$\endgroup$
$$i = \exp\left(2\pi n + \frac{\pi}2 \right)$$
$$i^\frac12 = \exp\left(\pi n + \frac{\pi}4 \right)=\cos\left(\pi n+\frac{\pi}{4}\right)+i\sin\left(\pi n +\frac{\pi}{4} \right)$$
answered Nov 16, 2017 at 15:51