Question

Find modulus of z = (1+i)^13/(1-i)^7

Answer 1

Modulus of z is 8

Step-by-step explanation:

let,

$z=\frac{(1+i)^{13}}{(1-i)^{7}}\\\\=\frac{((1+i)^{2})^{6}(1+i)}{((1-i)^{2})^{6}(1-i)}\\\\=\frac{((1+i)^{2})^{6}(1+i)}{((1-i)^{2})^{6}(1-i)} \times \frac{(1+i)}{(1+i)}$

$=\frac{(1+2i-1)^{6}(1+i)^{2}}{(1-2i-1)^{3}(1-i^{2})}\\\\=\frac{2^{6}i^{6}(1+2i+i^{2})}{-2^{3}i^{3}(1-i^{2})}\\\\=\frac{2^{3}i^{3}(2i)}{(1-(-1))}\\\\=-2^{3}i^{4}\\\\=-8$

$Thus, Modulus r=|z|=|-8|=8$

And,point (−8,0) representing$z=-8+0i$ lies on the negative side of real axis.

Therefore, Principal Argument of z is $\pi$

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Answer 2

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