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Mathematics Question on Complex Numbers and Quadratic Equations

Convert (i+1)(cos⁡ (π4)−i sin⁡(π4)\frac {(i+1)}{(\cos \,(\frac{\pi}{4}) -i\,\sin (\frac {\pi} {4})} in polar form?

A

cos⁡ (π/4)+i sin (π/4)\cos \,(\pi /4)+i\,sin\,(\pi /4)

B

cos⁡ (π/2)+i sin⁡ (π/2)\cos \,(\pi /2)+i\,\sin \,(\pi /2)

C

2 (cos⁡ (π/4)+i sin⁡(π/4))\sqrt{2}\,(\cos \,(\pi /4)+i\,\sin (\pi /4))

D

2 (cos⁡ (π/2) + i sin⁡(π/2))\sqrt{2}\,(\cos \,(\pi /2)\,+\,i\,\sin (\pi /2))

Answer

2 (cos⁡ (π/2) + i sin⁡(π/2))\sqrt{2}\,(\cos \,(\pi /2)\,+\,i\,\sin (\pi /2))

Explanation

Solution

i+1cos⁡π4−isin⁡π4=i+112(1−i)×1+i1+i\frac{i+1}{\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}}=\frac{i+1}{\frac{1}{\sqrt{2}(1-i)}}\times \frac{1+i}{1+i}
=2(1+i)21−i2=\frac{\sqrt{2}{{(1+i)}^{2}}}{1-{{i}^{2}}}
=2(1−1+2i)1+1=\frac{\sqrt{2}(1-1+2i)}{1+1}
=2.2i2=2i=\frac{\sqrt{2}.2i}{2}=\sqrt{2}i
=2(cos⁡ π2+i sin⁡π2)=\sqrt{2}\left( \cos \,\frac{\pi }{2}+i\,\sin \frac{\pi }{2} \right)

For a single coordinate point, the formula allows us to produce an endless number of polar coordinates. The equation is expressed as follows:

(r, θ + 2πn) or (-r, θ + (2n+1)π)

n is an integer in this situation.

The value of will be positive if measured in an anticlockwise direction and negative if measured in a clockwise direction. The value of r will also be positive if you lay off the terminal side, but negative if you lay off the prolongation via the origin from the terminal side. The starting side of an angle is referred to as such, while the terminal side is the ray on which the angle measurement terminates.