If |z|<2−1, then z2+2zcos⁡α is

If |z|

A
less than 1
B
$\sqrt{2} + 1$
C
$\sqrt{2} - 1$
D
none of these

Solution:

$\begin{array}{l} ∣ z^{2} + 2 zcos α | \leq | z^{2} | + | 2 zcos α ∣ \\ = | z |^{2} + 2 | z | | \cos α | \\ \leq | z |^{2} + 2 | z | (∵ | \cos α | \leq 1) \\ < (\sqrt{2} - 1)^{2} + 2 (\sqrt{2} - 1) = 1 \end{array}$

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