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Complex numbers help

Hey guyz,am used on normal problems with no fraction,but this has fraction.Any one who can helpFind the modulus and argument of the following;(i)z=5+i/2+3i (ii)z=(square root of 3+i / 1+i square root of 3)
Reply 1
Original post by shady2.0
Hey guyz,am used on normal problems with no fraction,but this has fraction.Any one who can helpFind the modulus and argument of the following;(i)z=5+i/2+3i (ii)z=(square root of 3+i / 1+i square root of 3)


If you like complex exponentials, then you can write the numerator and the denominator in the form reiθre^{i\theta} and then deduce both from there.

Alternatively, you can compute the modulus as for w,zCw,z \in \mathbb{C} we have that zw=zw\left| \dfrac{z}{w} \right| = \dfrac{|z|}{|w|}.
The argument is given by arg(zw)=arg(z)arg(w)\arg\left( \dfrac{z}{w} \right)=\arg(z)-\arg(w).
Reply 2
If
Unparseable latex formula:

\[z = \frac{5 + i}{2+3i}

, then how would you simplify z into one complex number in the form a + bi? Remember that the polar form of a complex number is
Unparseable latex formula:

z = r(\cos \theta + i\cdot \sin \theta )\]

.
Reply 3
Original post by aymanzayedmannan
If
Unparseable latex formula:

\[z = \frac{5 + i}{2+3i}

, then how would you simplify z into one complex number in the form a + bi? Remember that the polar form of a complex number is
Unparseable latex formula:

z = r(\cos \theta + i\cdot \sin \theta )\]

.
Thank you I got it now
Reply 4
Original post by joostan
If you like complex exponentials, then you can write the numerator and the denominator in the form reiθre^{i\theta} and then deduce both from there.

Alternatively, you can compute the modulus as for w,zCw,z \in \mathbb{C} we have that zw=zw\left| \dfrac{z}{w} \right| = \dfrac{|z|}{|w|}.
The argument is given by arg(zw)=arg(z)arg(w)\arg\left( \dfrac{z}{w} \right)=\arg(z)-\arg(w).
Thank you got it now

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