Question

find the modulus and argument of the complex number and hence express them in polar form 1+2i/1-(1-i)²

Answer 1

Dear student

$$\begin{gathered} \frac{1+2i}{1-(1-i{)}^2} \\ =\frac{1+2i}{1-(1+i^2-2i)} \\ =\frac{1+2i}{1-(1-1-2i)} \\ =\frac{1+2i}{1-(0-2i)} \\ =\frac{1+2i}{1+2i} \\ =1 \\ ifz=x+iy=1 \\ x=1 \\ y=0 \\ \left|z\right|=\sqrt{x^2+y^2}=1 \\ arg\left(z\right)={\tan }^{-1}(y/x)={\tan }^{-1}(0/1)={\tan }^{-1}\left(0\right)=0 \\ z=x+iy \\ =\left|z\right|(\cos \theta +i\sin \theta )inpoarform \\ =\left|1\right|(\cos 0+i\sin 0) \end{gathered}$$

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