Thanks everyone that answered my last question! I had one more question for a different concept. If the modulus of a complex number $z = |z| = \sqrt{(a^2 + b^2)}$,where $a$ and $b$ are the real and imaginary constants of $z$, are you allowed to use the similar property for conjugates and say that the modulus of $(z + 1)$, or $|z + 1| = |z| + |1|$ or would it instead be $\sqrt{(a^2 + b^2 + 1^2)}$?
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1I don't think it is any of those, I think it may be $|z+1|=\sqrt{(a+1)^2 + b^2}$ – Joshua Pasa Apr 19 '20 at 21:03
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1Since 1 is real – Joshua Pasa Apr 19 '20 at 21:03
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1Remember that: $|z_1 + z_2| \leq |z_1| + |z_2|$ by triangle inequality – tajiri_numero_1 Apr 19 '20 at 21:06
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$|z+1|=\sqrt{(z+1)(z'+1)}=\sqrt{zz'+z+z'+1}=\sqrt{a^2+b^2+2a+1}$ if $z=a+bi$.
Alternatively, $|z+1|=\sqrt{(z+1)(z'+1)}=\sqrt{(a+1+bi)(a+1-bi)}=\sqrt{(a+1)^2+b^2}$.
J. W. Tanner
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In general if $z=x+iy$, then $|z|=\sqrt{x^{2}+y^{2}}$ so that if $z+1=(x+1)+iy$ we have $|z|=\sqrt{(x+1)^{2}+y^{2}}$.
Alessio K
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