Theory of Intelligence
11TH•Mathematics•Complex Numbers•Question 2

(v) $\dfrac{1-i}{1+i}$

Express the following complex numbers in polar form:

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(v) 1−i1+i\dfrac{1-i}{1+i}

1−i1+i=1−i1+i⋅1−i1−i=(1−i)21−i2=1−2i+i22=−2i2=−i\begin{aligned} \frac{1-i}{1+i} &=\frac{1-i}{1+i}\cdot\frac{1-i}{1-i} =\frac{(1-i)^2}{1-i^2}\\ &=\frac{1-2i+i^2}{2}=\frac{-2i}{2}=-i \end{aligned}

So z=−iz=-i.

r=∣z∣=1,θ=−π2 r=|z|=1,\quad \theta=-\frac{\pi}{2} −i=cos⁡(−π2)+isin⁡(−π2)\boxed{-i=\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)}

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