Not so complex

Algebra Level 4

cis ( x ) = e π 2 i \large \text{cis}(x) = e^{\frac{\pi}{2}i}

Given that x x is a complex number in the form a + b i a + bi that follows the equation above. Find a + b a + b .

Details and Assumptions

  • cis ( x ) = cos ⁡ ( x ) + i sin ⁡ ( x ) \text{cis}(x) = \cos(x) + i\sin(x)
  • i i is the imaginary number, i = − 1 i = \sqrt{-1}


The answer is 0.

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Jack Rawlin by Jack Rawlin , 22, United Kingdom

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1 solution

Jack Rawlin
Mar 2, 2016

cis ( x ) = e π 2 i \large \text{cis}(x) = e^{\frac{\pi}{2}}i

cis ( a + b i ) = e π 2 e π 2 i \large \text{cis}(a + bi) = e^{\frac{\pi}{2}}e^{\frac{\pi}{2}i}

e ( a + b i ) i = e π 2 + π 2 i \large e^{(a + bi)i} = e^{\frac{\pi}{2} + \frac{\pi}{2}i}

e − b + a i = e π 2 + π 2 i \large e^{-b + ai} = e^{\frac{\pi}{2} + \frac{\pi}{2}i}

− b + a i = π 2 + π 2 i \large -b + ai = \frac{\pi}{2} + \frac{\pi}{2}i

− b = π 2 , a i = π 2 i \large -b = \frac{\pi}{2},~ ai = \frac{\pi}{2}i

b = − π 2 , a = π 2 \large b = -\frac{\pi}{2},~ a = \frac{\pi}{2}

a + b = π 2 + ( − π 2 ) \large a + b = \frac{\pi}{2} + \left(-\frac{\pi}{2}\right)

a + b = π 2 − π 2 \large a + b = \frac{\pi}{2} - \frac{\pi}{2}

a + b = 0 \large \boxed{a + b = 0}

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