Question 526946
Since {{{-10-5i}}} is in standard form {{{a+bi}}}, we can see that {{{a=-10}}} and {{{b=-5}}}



{{{abs(a+bi)=sqrt(a^2+b^2)}}} Start with the absolute value of a complex number formula.



{{{abs(-10-5i)=sqrt((-10)^2+(-5)^2)}}} Plug in {{{a=-10}}} and {{{b=-5}}}.



{{{abs(-10-5i)=sqrt(100+(-5)^2)}}} Square {{{-10}}} to get {{{100}}}.



{{{abs(-10-5i)=sqrt(100+25)}}} Square {{{-5}}} to get {{{25}}}.



{{{abs(-10-5i)=sqrt(125)}}} Add.



{{{abs(-10-5i)=5*sqrt(5)}}} Simplify the square root.



So the answer is {{{abs(-10-5i)=5*sqrt(5)}}} 



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim