জটিল সংখ্যা ও জ্যামিতিক প্রতিরূপ
3+2i3−i=a+ib \frac{3 + 2 i}{3 - i} = a + i b 3−i3+2i=a+ib হলে b = কত?
910 \frac{9}{10} 109
3+2i3−i=(3+2i)(3+i)(3−i)(3+i) \begin{aligned} & \frac{3+2 i}{3-i} \\ = & \frac{(3+2 i)(3+i)}{(3-i)(3+i)}\end{aligned} =3−i3+2i(3−i)(3+i)(3+2i)(3+i)
=9+6i+3i−29−i2=7+9i10=710+910i=a+bi. \begin{array}{l}=\frac{9+6 i+3 i-2}{9-i^{2}} \\ =\frac{7+9 i}{10} \\ =\frac{7}{10}+\frac{9}{10} i \\ =a+b i .\end{array} =9−i29+6i+3i−2=107+9i=107+109i=a+bi.
∴b=910 \therefore b=\frac{9}{10} ∴b=109
The roots of ax2+bx+c=0a{x^2} + bx + c = 0ax2+bx+c=0 ,where a≠0a \ne 0a=0 ,b,c are non-real complex and
a+c<ba + c < ba+c<b , then
If (x2−2)+(y+3)i=7+4i(x^{2}-2)+(y+3)i=7+4i(x2−2)+(y+3)i=7+4i then x and y are
Given that i=−1i = \sqrt {-1}i=−1, find the multiplicative inverse of 5−i5 - i5−i.
y এবং z এককের ঘনমূল হলে-
i. Z5=yZ^5=yZ5=y
ii. z7+y7=i6z^7+y^7=i^6z7+y7=i6
iii. z2y2=i4z^2y^2=i^4z2y2=i4
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