Riddle Me This

Using $\space i^2+1=0$ $\sqrt{2i}=\sqrt{i^2+2i+1}=\sqrt{(i+1)^2}= i+1$ $\Large a=b=1\space \text{or} \space a+b=2$

When rewriting a complex number in an alternate form, it is always helpful to first rewrite it as $z=e^{\ln (z) }$ . Executing:

$\sqrt{2i}= (2i)^{\frac{1}{2}} = e^{\frac{1}{2} \ln (2i)}=e^{\frac{1}{2} ( \ln 2 + \ln i )}=e^{\ln \sqrt{2}+\frac{1}{2} \ln (i)} = \exp \left ( \ln \sqrt{2}+\frac{i \pi}{4} \right )$

Using the definition of the complex logarithm $\textrm{Log} (z) = \ln |z| +i \textrm{Arg} (z)$ we can see that $\ln |z|= \ln \sqrt{2}$ and $i \textrm{Arg} (z) =\frac{i \pi}{4}$ . So, we need to come up with some complex number with a modulus of $\sqrt{2}$ and an argument of $\frac{i \pi}{4}$ . We know that the argument of $z$ is $|a+ib|=\sqrt{a^2+b^2}$ so let's generate some possible values of $a$ and $b$ that will sum to 2. Well, if $a=1$ and $b=1$ then $a^2+b^2=2$ . Simple. However, we don't know if these numbers will generate the argument required. let's check.

Let's picture our potential complex number as a vector in the complex plane. We have an $\hat{\imath}$ -component (real part) of 1 and a $\hat{\jmath}$ -component (imaginary part) of 1. Using the definition of tangent we see that

$\tan \theta =\frac{\textrm{opposite}}{\textrm{adjacent}} \: \Rightarrow \: \tan \theta =1 \: \rightarrow \: \theta= \frac{\pi}{4}$

So everything checks out and our guess is correct! Therefore $a=b=1$ and so $\sqrt{2i}=1+i$

$a+b=\Re [ a+ib ] + \Im [ a+ib ] = \boxed{2}$

$\sqrt{2i} = \sqrt{2e^{\frac{i\pi}{2}}} = \sqrt{2} \cdot e^{\frac{i\pi}{4}} = \sqrt{2}\left (\cos \frac{\pi}{4} +i\sin \frac{\pi}{4} \right) =$ = $\sqrt{2} \left( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right) = 1 + i \Rightarrow a = b= 1 \Rightarrow a + b = \boxed{2}$

$\sqrt{2i} = \sqrt{2cis(\frac{\pi}{2})}$ $\sqrt2 \cdot (\frac{1}{\sqrt2}+\frac{i}{\sqrt2})$ $1+i$ $1+1=2$