Question 1178611


Solve the following for {{{z}}}: 

{{{z^2=5-12i}}}

substitute {{{z=x+yi}}}

{{{(x+yi)^2=5-12i}}}................expand

{{{x^2+2yi+(yi)^2=5-12i}}}

{{{x^2+2yi+y^2(i)^2=5-12i}}}

{{{x^2+2yi+y^2(-1)=5-12i}}}

{{{x^2+2yi-y^2=5-12i}}}

{{{(x^2-y^2) +2yi=5-12i}}}

complex numbers can be equal only if their real and imaginary parts are equal

so,

{{{x^2-y^2=5}}}=>{{{(x - y) (x + y) = 5}}}: integer solutions that satisfy this are: {{{x}}} = ± {{{3}}}, {{{y}}} = ±{{{ 2}}}

{{{2yi=-12i}}}=> {{{yi=-6i}}}

substitute back {{{z=x+yi}}}

if {{{x=3}}} and {{{y=2}}}

{{{z=3+2i}}}-> also {{{z=3-2i}}}

or

if {{{x=-3}}} and {{{y=2}}}

{{{z = -3 + 2 i}}}-> also {{{z = -3 - 2 i}}}