How do I use the modulus and argument to square #(1+i)#?

1 Answer
Daniel L.
Apr 27, 2018

You should use the de Moivre's Theorem. See explanation.

Explanation:

For any complex number #z# given in trigonometric form (having modulus #|z|# and argument #varphi#) we can calculate any natural power using the formula:

#z^n=|z|^n*(cosnvarphi+i*sinnvarphi)#

Here we get:

#|z|=sqrt(1^2+1^2)=sqrt(2)#

#cosvarphi=(Re(z))/|z|=1/sqrt(2)=sqrt(2)/2=> varphi=45^o#

Now we can calculate the square:

#z^2=|z|^2*(cos(2*45)+isin(2*45))=#

#=2*(cos90+isin90)=2*(0+1i)=2i#

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