Express in the modulus and amplitude form- 1- i 2 n -1-1- i 22-1- where n belongs to natural number-

(1+i)^2n+1/(1 i)^2n 1 Multiplying both numerator and denominater by (1+1)^2n 1 (1+i)^2n+1(1 i)^2n 1/(1 i)^2n 1(1+i)^2n 1 (1+i)^2n+1/{ (1 i)(1_i) }^2n 1 { (1+i)^ ...

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Express in the modulus and amplitude form. $\frac{(1 + i)^{2 n + 1}}{(1 - i)^{2 n - 1}}$ , where n belongs to natural number.

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Solution

$\frac{(1 + i)^{2 n + 1}}{(1 - i)^{2 n - 1}}$
Multiplying both numerator and denominater by $(1 + 1)^{2 n - 1}$
$\frac{(1 + i)^{2 n + 1} (1 - i)^{2 n - 1}}{(1 - i)^{2 n - 1} (1 + i)^{2 n - 1}} \frac{(1 + i)^{2 n + 1}}{{(1 - i) (1_{i})}_{i}^{2 n - 1}} \frac{{(1 + i)^{2}}^{2 n}}{(2)^{2 n - 1}} \frac{{1 + i^{2} + 2 i}^{2 n}}{(2)^{2 n - 1}} [∵ i^{2} = - 1 s o i^{2} + 1 = 0] \frac{(2 i)^{2 n}}{(2)^{2 n - 1}} \frac{2^{2 n} (i^{2})^{n}}{2^{2 n - 1}} \Rightarrow 2 \times (- 1)^{n} \to 2 \times (- 2)^{n}$

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Express in the modulus and amplitude form.(1+i)2n+1(1−i)2n−1, where n belongs to natural number.

(1+i)2n+1(1−i)2n−1 Multiplying both numerator and denominater by (1+1)2n−1 (1+i)2n+1(1−i)2n−1(1−i)2n−1(1+i)2n−1(1+i)2n+1{(1−i)(1i)}2n−1{(1+i)2}2n(2)2n−1{1+i2+2i}2n(2)2n−1[∵i2=−1 so i2+1=0](2i)2n(2)2n−122n(i2)n22n−1⇒2×(−1)n→2×(−2)n

Express in the modulus and amplitude form. $\frac{(1 + i)^{2 n + 1}}{(1 - i)^{2 n - 1}}$ , where n belongs to natural number.