Question 1196157
<pre>

Did you mean this: {{{sqrt(1+2i)/(1-2i)}}} or this: {{{sqrt((1+2i)/(1-2i))}}}?

I'll assume the second

{{{sqrt((1+2i)/(1-2i))}}}

{{{sqrt(((1+2i)(1+2i))/((1-2i)(1+2i)))}}}

{{{sqrt((1+4i+4i^2)/(1-4i^2))}}}

{{{sqrt((1+4i+4(-1))/(1-4(-1)))}}}

{{{sqrt((1+4i-4)/(1+4))}}}

{{{sqrt((-3+4i)/(5))}}}

{{{sqrt(1/5)sqrt(-3+4i)}}}

{{{sqrt(expr(1/5)*expr(5/5))sqrt(-3+4i)}}}

{{{expr(sqrt(5)/5)sqrt(-3+4i)}}}

let

{{{sqrt(-3+4i)}}}{{{""=""}}}{{{a+bi}}}

{{{-3+4i}}}{{{""=""}}}{{{a^2+2abi+b^2i^2}}}

{{{-3+4i}}}{{{""=""}}}{{{a^2+2abi+b^2(-1)}}}

{{{-3+4i}}}{{{""=""}}}{{{a^2+2abi-b^2}}}

Setting real parts equal and imaginary parts equal.

{{{system(a^2-b^2=-3, 2ab=4)}}}

{{{system(a^2-b^2=-3, ab=2)}}}

Solve that system and get a = ±1, b = ±2

So there are two solutions to the system: ±1±2i

In either case the modulus is {{{(""+-1)^2+(""+-2)^2}}}{{{""=""}}}{{{sqrt(1+4)}}}{{{""=""}}}{{{sqrt(5)}}}

But the modulus must be multiplied by the factor {{{sqrt(5)/5}}}

{{{modulus}}}{{{""=""}}}{{{sqrt(5)/5}}}{{{""*""}}}{{{sqrt(5)}}}}}}""=""}}}{{{5/5}}}{{{""=""}}}{{{1}}}

In the case of 1+2i, which is in the first quadrant with 

{{{tan(argument)}}}{{{""=""}}}{{{2/1}}}{{{""=""}}}{{{2}}}

{{{argument}}}{{{""=""}}}{{{63.43^o}}}

In the case of -1-2i, which is in the third quadrant with 

{{{tan(argument)}}}{{{""=""}}}{{{(-2)/(-1)}}}{{{""=""}}}{{{2}}}

{{{argument}}}{{{""=""}}}{{{243.43^o}}}

Edwin</pre>