Question

Find the modulus and amplitude for each of the following complex numbers.
i) $7-5i$ vi) $\sqrt{3}-i$
ii) $\sqrt{3}+\sqrt{2i}$ vii) $3$
iii) $-8+15i$ viii) $1+i$
iv) $-3(l-i)$ ix) $1+i\sqrt{3}$
v) $-4-4i$ x) $(1+2i{)}^2(1-i)$

Answer 1

Complex number: $z=x+iy$

Modulus: $z=\sqrt{x^2+y^2}$

Amplitude: $\theta ={\tan }^{-1}\left|\frac{b}{a}\right|$

(i)

$z=7-5i$

$\Rightarrow \left|z\right|=\sqrt{7^2+(-5{)}^2}=8.6$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{-5}{7}\right|=35.53$

(ii)

$z=\sqrt{3}+\sqrt{2}i$

$\Rightarrow \left|z\right|=\sqrt{{\sqrt{3}}^2+{\sqrt{2}}^2}=2.23$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{\sqrt{2}}{\sqrt{3}}\right|=39.23$

(iii)

$z=-8+15i$

$\Rightarrow \left|z\right|=\sqrt{(-8{)}^2+{15}^2}=17$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{\left(15\right)}{-8}\right|=61.92$

(iv)

$z=-3+3i$

$\Rightarrow \left|z\right|=\sqrt{(-3{)}^2+3^2}=4.24$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{(-3)}{3}\right|=45$

(v)

$z=-4-4i$

$\Rightarrow \left|z\right|=\sqrt{(-4{)}^2+(-4{)}^2}=5.65$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{(-4)}{(-4)}\right|=45$

(vi)

$z=\sqrt{3}-i$

$\Rightarrow \left|z\right|=\sqrt{(\sqrt{3}{)}^2+(-1{)}^2}=2$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{(-1)}{\left(\sqrt{3}\right)}\right|=30$

(vii)

$z=3+i\left(0\right)$

$\Rightarrow \left|z\right|=\sqrt{(3{)}^2+(0{)}^2}=3$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{\left(0\right)}{\left(3\right)}\right|=0$

(viii)

$z=1+i$

$\Rightarrow \left|z\right|=\sqrt{(1{)}^2+(1{)}^2}=1.414$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{\left(1\right)}{\left(1\right)}\right|=45$

(ix)

$z=1+\sqrt{3}i$

$\Rightarrow \left|z\right|=\sqrt{(1{)}^2+(\sqrt{3}{)}^2}=2$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{\left(\sqrt{3}\right)}{\left(1\right)}\right|=60$

(x)

$(1+2i{)}^2(1-i)=(-3+4i\left)\right(1-i)=-6+7i$

$z=-6+7i$

$\Rightarrow \left|z\right|=\sqrt{(-6{)}^2+(7{)}^2}=2$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{\left(\sqrt{7}\right)}{(-6)}\right|=49.39$