To be able to write some numbers in complex form, a little algebraic manipulation may be necessary, mostly involving the fact that $\sqrt{1}=i$√−1=i or that $1=i^2$−1=i2. Let's have a look at some examples.
Rewrite $\sqrt{50}$√−50 in the form $z=x+yi$z=x+yi
$\sqrt{50}$√−50  $=$=  $\sqrt{1\times50}$√−1×50 
_{using properties of surds} $\sqrt{ab}=\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$√ab=√a×b=√a×√b 
$=$=  $\sqrt{1}\times\sqrt{50}$√−1×√50  
$=$=  $\sqrt{50}i$√50i 
_{and now simplify the }$\sqrt{50}$√50 $\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt{2}=5\sqrt{2}$√50=√25×2=√25×√2=5√2 

$=$=  $5\sqrt{2}i$5√2i 
So in the form $x+yi$x+yi, the real component is $0$0, ie. that $x=0$x=0 and the imaginary component is $y=5\sqrt{2}$y=5√2, so $x+yi=0+5\sqrt{2}i$x+yi=0+5√2i
Consider first the fact that a square root sign is actually an index. It is the power half.
$\sqrt{a}=a^{\frac{1}{2}}$√a=a12
Now, also remember all the index laws.
$\left(ab\right)^n$(ab)n  $=$=  $a^n\times b^n$an×bn  (1) 
$\left(\frac{a}{b}\right)^n$(ab)n  $=$=  $\frac{a^n}{b^n}$anbn  (2) 
We can now look at what these rules look like with regards to square roots.
Let's start here with (1) $\left(ab\right)^n=a^n\times b^n$(ab)n=an×bn with $n=\frac{1}{2}$n=12
$\left(ab\right)^{\frac{1}{2}}$(ab)12  $=$=  $a^{\frac{1}{2}}\times b^{\frac{1}{2}}$a12×b12  which is equivalent to 
$\sqrt{ab}$√ab  $=$=  $\sqrt{a}\times\sqrt{b}$√a×√b 
The other equation (2) $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$(ab)n=anbn with $n=\frac{1}{2}$n=12 becomes
$\left(\frac{a}{b}\right)^{\frac{1}{2}}$(ab)12  $=$=  $\frac{a^{\frac{1}{2}}}{b^{\frac{1}{2}}}$a12b12  which becomes 
$\sqrt{\frac{a}{b}}$√ab  $=$=  $\frac{\sqrt{a}}{\sqrt{b}}$√a√b 
$$ which becomes $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$√ab=√a√b
This covers multiplication and division with square roots.
Multiplication of Square Roots $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$√ab=√a×√b
Division of Square Roots $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$√ab=√a√b
Now, what about addition and subtraction I hear you ask. Well, there are no simplification laws for general addition and subtraction of square roots, unless the value in the radicand is the same.
Addition and subtraction only works if the value in the square root is identical.
Thus
$a\sqrt{c}+b\sqrt{c}=\left(a+b\right)\sqrt{c}$a√c+b√c=(a+b)√c
and
$a\sqrt{c}b\sqrt{c}=\left(ab\right)\sqrt{c}$a√c−b√c=(a−b)√c
But $\sqrt{a}+\sqrt{b}$√a+√b is definitely NOT equal to $\sqrt{a+b}$√a+b
Combining together now our ability to convert $i^2=\sqrt{1}$i2=√−1, and our knowledge of the properties of surds we can simplify a whole myriad of questions involving negative surds.
Things to remember when doing this
Here is an example of what I mean by that last point
Simplify $\sqrt{2}\times\sqrt{30}$√−2×√−30
$\sqrt{2}\times\sqrt{30}$√−2×√−30  $=$=  $\sqrt{1\times2}\times\sqrt{1\times30}$√−1×2×√−1×30 
$=$=  $i\sqrt{2}\times i\sqrt{30}$i√2×i√30  
$=$=  $i^2\sqrt{2\times30}$i2√2×30  
$=$=  $\sqrt{4\times15}$−√4×15  
$=$=  $2\sqrt{15}$−2√15 
A very common error for students to make here is to assume that the negative 2 multiplied by the negative 30 would give positive 60 and hence they would end up with the answer of $2\sqrt{15}$2√15. So deal with each $\sqrt{1}$√−1 component separately.
Express $\sqrt{100}$√−100 in terms of $i$i.
Express $\sqrt{29}$−√−29 in terms of $i$i.
Find the value of $\frac{\sqrt{33}\times\sqrt{3}}{\sqrt{11}}$√−33×√−3√11.
Manipulate complex numbers and present them graphically
Apply the algebra of complex numbers in solving problems