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Question Number 112364 by john santu last updated on 07/Sep/20

(1) L e t z = \sqrt{3 + 4 i} + \sqrt{- 3 - 4 i}, w h e r e

s q u a r e r o o t i s t a k e n w i t h p o s i t i v e

r e a l p a r t . T h e n R e (z) i s_

(a) 3 (b) 4 (c) 2 (d) 1

(2) S u p p o s e a, b, c a r e i n A P a n d a^{2}, b^{2}, c^{2}

a r e i n G P . I f a < b < c a n d a + b + c = \frac{3}{2},

t h e n a =_

(a) \frac{1}{2 \sqrt{2}} (b) \frac{1}{2 \sqrt{3}} (c) \frac{\sqrt{3} - 2}{2 \sqrt{3}} (d) \frac{\sqrt{2} - 2}{2 \sqrt{2}}

(3) A m a n s e n t 7 l e t t e r s t o h i s 7 f r i e n d

. T h e l e t t e r s a r e k e p t i n a d d r e s s e d

e n v e l o p e s a t r a n d o m . T h e p r o b a b i l i t y

t h a t 3 f r i e n d r e c e i v e c o r r e c t l e t t e r s

a n d 4 l e t t e r s g o t o w r o n g d e s t i n a t i o n

i s_

(a) \frac{1}{8} (b) \frac{1}{16} (c) \frac{3}{32} (d) \frac{5}{64}

Answered by MJS_new last updated on 07/Sep/20

\sqrt{z} is unique

\sqrt{3 + 4 i} = 2 + i

\sqrt{- 3 - 4 i} = 1 - 2 i

\Rightarrow z = 3 - i

Commented by MJS_new last updated on 07/Sep/20

x^{2} = 4

{we}^{'} re looking for a l l p o s s i b l e numbers x which

solve the equation \Rightarrow x = \pm 2

2^{2} we just square without caring if or if not

{(any other number)}^{2} = 2^{2} = 4

\sqrt{x} = 2

{we}^{'} re looking for a l l p o s s i b l e numbers x which

solve the equation \Rightarrow x = 4 (try to find another

solution x \in C)

\sqrt{4} we just extract the root without caring if

or if not \sqrt{any other number} = \sqrt{4} = 2

so {there}^{'} s no equation given, nothing to

solve when we just calculate 7^{2} or \sqrt{49}

btw . {there}^{'} s no solution to \sqrt{x} = - 4 (try it)

z = r e^{i θ} with r ⩾ 0 and θ \in R

z^{q} = r^{q} e^{i q θ} nothing to solve with z, q given

{there}^{'} s an exception made to keep geometric

problems real :

z^{\frac{1}{n}} with z < 0 and n = 2 k + 1 \land k \in Z

\Rightarrow z = - {(- z)}^{\frac{1}{n}}

or z = r e^{i π} \Rightarrow z^{\frac{1}{2 k + 1}} = r^{\frac{1}{2 k + 1}} e^{i π} n o t r^{\frac{1}{2 k + 1}} e^{\frac{i π}{2 k + 1}}

Commented by john santu last updated on 08/Sep/20

g r e a t t