Question

Convert,the complex number into polar form: 4+4i

Answer 1

Z = a + bi
z = 4 + 4i

r² = a² + b²
r² = (4)² + (4)²
r² = 16 + 16
r² = 32
 r = 4√(2)
 r = 4(1.414)
 r = 5.656

$cos\theta = \frac{a}{r}$
$cos\theta = \frac{4}{4\sqrt{2}}$
$cos\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$cos\theta = \frac{4\sqrt{2}}{4\sqrt{4}}$
$cos\theta = \frac{4\sqrt{2}}{4(2)}$
$cos\theta = \frac{4\sqrt{2}}{8}$
$cos\theta = \frac{\sqrt{2}}{2}$
$2(cos\theta) = 2(\frac{\sqrt{2}}{2})$
$2cos\theta = \sqrt{2}$
$2cos\theta = 1.414$

$sin\theta = \frac{b}{r}$
$sin\theta = \frac{4}{4\sqrt{2}}$
$sin\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$sin\theta = \frac{4\sqrt{2}}{4\sqrt{4}}$
$sin\theta = \frac{4\sqrt{2}}{4(2)}$
$sin\theta = \frac{4\sqrt{2}}{8}$
$sin\theta = \frac{\sqrt{2}}{2}$
$2(sin\theta) = 2(\frac{\sqrt{2}}{2})$
$2sin\theta = \sqrt{2}$
$2sin\theta = 1.414$

z = a + bi
z = rcosθ + (rsinθ)i
z = r(cosθ + i sinθ)

z = 4 + 4i
z = 5.656cosθ + (5.656sinθ)i
z = 5.656(cosθ + i sinθ)
z = 5.656(cos45 + i sin45)

$\theta = tan^{-1}\frac{b}{a}$
$\theta = tan^{-1}\frac{4}{4}$
$\theta = tan^{-1}(1)$
$\theta = 45$

The polar form of 4 + 4i is approximately equal to 5.656(cos45 + i sin45).