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Sia ? 6 years, 4 months ago
Let P (n) =(2n + 7) < (n + 3)2
For n = 1
{tex}P(1) = (2 \times 1 + 7) < {(1 + 3)^2} \Rightarrow 9 < 16{/tex}
{tex}\therefore {/tex} P ( 1) is true
Let P(n) be true for n = k
{tex}\therefore P(k) = (2k + 7) < {(k + 3)^2}{/tex} ....(1)
For n = k + 1
P (k + 1) = 2 (k + 1) + 7 < (k + 1 + 3)2
{tex} \Rightarrow {/tex} 2(k + 1) + 7 < (k + 4)2
From (1)
2k + 7 < (k + 3)2
Adding 2 on both sides
2k + 7 + 2 < (k + 3)2 + 2 {tex} \Rightarrow {/tex} 2(k + 1) + 7 < k2 + 9 + 6k + 2
{tex} \Rightarrow {/tex} 2(k + 1) + 7 < k2 +6k + 11 < k2 + 8k + 16
2(k + 1) + 7 <(k + 4)2
{tex}\therefore {/tex} P (k + 1) is true
Thus P(k) is true {tex} \Rightarrow {/tex} P(k + 1) is true
Hence by principle of mathematical induction, P (n) is true for all {tex}n \in N{/tex}.
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Sia ? 6 years, 4 months ago
Let {tex}x + yi = \sqrt { - i} {/tex}
Squaring both sides, we get
(x + yi)2 = -i
x2 - y2 + 2xyi = -i
Equating the real and imaginary parts
x2 - y2 = 0 . . . (i)
and 2xy = -1
{tex}\therefore xy = - \frac{1}{2}{/tex}
Now using the identity
(x2 + y2)2 = (x2 - y2)2 + 4x2y2
{tex} = {(0)^2} + 4{\left( { - \frac{1}{2}} \right)^2}{/tex}
= 1
{tex}\therefore {/tex} x2 + y2 = 1 . . . (ii) [Neglecting (-) sign as x2 + y2 > 0]
Solving (i) and (ii), we get
{tex}{x^2} = \frac{1}{2}{/tex} and {tex}{y^2} = \frac{1}{2}{/tex}
{tex}\therefore x = \pm \frac{1}{{\sqrt 2 }}{/tex} and {tex}y = \pm \frac{1}{{\sqrt 2 }}{/tex}
Since the sign of xy is (-) then if
{tex}x = \frac{1}{{\sqrt 2 }}{/tex}, {tex}y = - \frac{1}{{\sqrt 2 }}{/tex}
If {tex}x = - \frac{1}{{\sqrt 2 }}\;y = \frac{1}{{\sqrt 2 }}{/tex}
{tex}\style{font-family:Tahoma}{\style{font-size:8px}{\therefore\sqrt{-i}=\pm\left(\frac1{\sqrt2}-\frac1{\sqrt2}i\right)}}{/tex}
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